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Maths Solve the Following Question.(5 Marks)

Probability · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 98 questions.

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Solve the Following Question.(5 Marks)

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Question 15 Marks
Let $d_1, d_2, d_3$ be three mutually exclusive diseases. Let $S$ be the set of observable symptoms of these diseases. $A$ doctor has the following information from a random sample of 5000 patients: 1800 had disease $\mathrm{d}_1, 2100$ has disease $\mathrm{d}_2$, and others had disease $\mathrm{d}_3 .1500$ patients with disease $\mathrm{d}_1, 1200$ patients with disease $\mathrm{d}_2$, and 900 patients with disease $d_3$ showed the symptom. Which of the diseases is the patient most likely to have?
Answer
Let $E_1, E_2, E_3$ and A be events as:
$E_1$= Patient has disease $d_1$
$E_2$ = Patient has disease $d_2$
$E_3 $= Patient has disease $d_3$
A = Selected patient has symptom S.
$\text{P}(\text{E}_1)=\frac{1800}{5000}=\frac{18}{50}$
$\text{P}(\text{E}_2)=\frac{2100}{5000}=\frac{21}{50}$
$\text{P}(\text{E}_3)=\frac{1100}{5000}=\frac{11}{50}$
$P(A|E_1)$ = P(Patient with disease $d_1$ and shows symptom S)
$=\frac{1500}{1800}$
$=\frac{5}{6}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Patient with disease $d_2$ and symprom S)
$=\frac{1200}{2100}$
$=\frac{4}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Patient with disease $d_3​​​​​​​$ and symptom S)
$=\frac{900}{1100}$
$=\frac{9}{11}$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{5}{6}\times\frac{18}{50}}{\frac{5}{6}\times\frac{18}{50}+\frac{21}{50}\times\frac{4}{7}+\frac{11}{50}\times\frac{9}{11}}$
$=\frac{\frac{3}{10}}{\frac{3}{10}+\frac{6}{25}+\frac{9}{50}}$
$=\frac{3}{10}\times\frac{50}{36}$
$=\frac{5}{12}$
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{21}{50}\times\frac{4}{7}}{\frac{5}{6}\times\frac{18}{50}+\frac{21}{50}\times\frac{4}{7}+\frac{11}{50}\times\frac{9}{11}}$
$=\frac{\frac{6}{25}}{\frac{3}{10}\times\frac{6}{25}+\frac{9}{50}}$
$=\frac{6}{25}\times\frac{50}{36}$
$=\frac{1}{3}$
$\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{11}{50}\times\frac{9}{11}}{\frac{5}{6}\times\frac{18}{50}+\frac{21}{50}\times\frac{4}{7}+\frac{11}{50}\times\frac{9}{11}}$
$=\frac{\frac{9}{50}}{\frac{3}{10}+\frac{6}{25}+\frac{9}{50}}$
$=\frac{9}{50}\times\frac{50}{36}$
So, probabilities of $d_1, d_2, d_3$ disease are $\frac{5}{12},\frac{1}{3},\frac{1}{4}$ respectively.
Hence, the patient is most likely to have $d_1$ diseased.
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Question 25 Marks
If A and B are two independent events such that $\text{P}(\overline{\text{A}}\cap\text{B})=\frac{2}{15}$ and $\text{P}(\text{A}\cap\overline{\text{B}})=\frac{1}{6}$, then find P(B).
Answer
We are given
$\text{P}(\overline{\text{A}}\cap\text{B})=\frac{2}{15}$
$\text{P}(\text{A}\cap\overline{\text{B}})=\frac{1}{6}$
Since A, B are independent,
$\therefore\ \text{P}(\overline{\text{A}})\text{ P(B)}=\frac{2}{15}\Rightarrow\ [1-\text{P(A)}]\text{P(B)}=\frac{2}{15}\ .....(\text{i})$
and $\text{P(A) }\text{P }(\overline{\text{B}})=\frac{1}{6}\Rightarrow \text{P(A)}[1-\text{P(B)}]=\frac{1}{6}\ .....\text{(ii)}$
From (i) we get
$\text{P(B)}=\frac{2}{15}\times\frac{1}{1-\text{P(A)}}$
Substituting this value in equation (ii) we get,
$\text{P(A)}\Big[1-\frac{1}{15(1-\text{P(A)})}\Big]=\frac{1}{6}$
$\Rightarrow\ \text{P(A)}\Big[\frac{15(1-\text{P(A)})-2}{15(1-\text{P(A)})}\Big]=\frac{1}{6}$
$\Rightarrow 6P(A) (13 - 15P(A)) = 15(1 - P(A))$
$\Rightarrow 2P(A) (13 - 15P(A)) = 5 - 5P(A)$
$\Rightarrow 26P(A) - 30[P (A)]^2 + 5P(A) - 5 = 0$
$\Rightarrow -30[P(A)]^2 + 31P(A) - 5 = 0$
This is a quadrati equation in $x = P(A)$ given as
$-30x^2 + 31x - 5 = 0$
$\Rightarrow 30x^2 - 31x + 5 = 0$
$\therefore\ \text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
Where, $a = +30, b = -31, C = +5$
$\Rightarrow\ \text{x}=\frac{31\pm\sqrt{(-31)^2-4(30)(5)}}{60}$
$=\frac{31\pm\sqrt{961-600}}{60}$
$=\frac{31\pm19}{60}$
$=\frac{50}{60},\frac{12}{60}$
$=\frac{5}{6},\frac{1}{5}$
$\therefore\ \text{P(A)}=\frac{5}{6}\text{ or }\frac{1}{5}$
Now,
$\text{P(A)}[1-\text{P(B)}]=\frac{1}{6}$
Putting $\text{P(A)}=\frac{5}{6}$
$\frac{5}{6}[1-\text{P(B)}]=\frac{1}{6}$
$1-\text{P(B)}=\frac{1}{5}$
$\text{P(B)}=1-\frac{1}{5}$
$\text{P(B)}=\frac{4}{5}$
Putting $\text{P(A)}=\frac{1}{5}$
$\frac{1}{5}[1-\text{P(B)}]=\frac{1}{6}$
$1-\text{P(B)}=\frac{5}{6}$
$\text{P(B)}=1-\frac{5}{6}$
$\text{P(B)}=\frac{1}{6}$
Hence $\text{P(B)}=\frac{4}{5} \text{ or }\frac{1}{6}$
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Question 35 Marks
A and B throw a pair of dice alternately. A wins the game if he gets a total of $7$ and B wins the game if he gets a total of $10$. If A starts the game, then find the probability that B wins.
Answer
Total of $7$ on the dice can be obtained in the following ways:
$(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$
Probability of getting a total of $7=\frac{6}{36}=\frac{1}{6}$
Probobility of not getting a total of $7=1-\frac{1}{6}=\frac{5}{6}$
Total of $10$ on the dice can be obtainced in the following ways:
$(4, 6), (6, 4), (5, 5)$
Probability of getting a total of $10=\frac{3}{36}=\frac{1}{12}$
Probability of not getting a total of $10=1-\frac{1}{12}=\frac{11}{12}$
Let E and F be the two events, defined as follows:
E = getting a total of $7$ in a single throw of a die
F= Getting a total of $10$ in a single throw of a dice
$\text{P(E)}=\frac{1}{6},\text{P}(\overline{\text{E}})=\frac{5}{6},\text{P(F)}=\frac{1}{12},\text{P}(\overline{\text{F}})=\frac{11}{12}$
A wins if he gets a total of $7$ in $1^{st}, 3^{rd}$ or $5^{th}$ ..... throws.
Probability of A getting a total of $7$ in the $1^{st}$ throw $=\frac{1}{6}$
A will get the $3^{rd}$ throw if he fails in the $1^{st}$ throw and B fails in hte $2^{nd}$​​​​​​​ throw.
Probability of A getting a total of $7$ in the $3^{rd}​​​​​​​$​​​​​​​ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{F}})\text{P(E)}=\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}$
Similarly, probability if getting a total of $7$ in the $5^{th}​​​​​​​$​​​​​​​ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{F}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{F}})\text{P(E)}=\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}$
ans so on
Probability of winning of A $=\frac{1}{6}+\Big(\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}\Big)+\Big(\frac{5}{6}\times\frac{11}{12}\times\frac{5}{6}\times\frac{11}{12}\times\frac{1}{6}\Big)+\ .....$
$=\frac{\frac{1}{6}}{1-\frac{5}{6}\times\frac{11}{12}}=\frac{12}{17}$
$\therefore$ Probability of winning of B $= 1$ - Probability of winning of A $=1-\frac{12}{17}=\frac{5}{17}$
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Question 45 Marks
$A, B$ and $C$ in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely?
Answer
Let E be event of getting a head.
$\text{P(E)}=\frac{1}{2}\Rightarrow\ \text{P}(\overline{\text{E}})=\frac{1}{2}$
If A stars the game,
$\Rightarrow$ A wins the game in $1^{th}, 4^{th}, 7^{th},$ ..... toss of coin.
P (A wins)
$=\text{P}(\text{E}\cup\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\text{E}\cup\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\text{E}\cup\ .....)$
$=\text{P}(\text{E})+\text{P}(\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\text{E})+\text{P}(\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\text{E})+\ .....$
$=\text{P}(\text{E})+\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}}) \text{P}(\overline{\text{E}}) \text{P}(\text{E})+\text{P}(\overline{\text{E}}) \text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\text{E})+\ .....$
$=\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\ .....$
$=\frac{1}{2}+\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^7+\ .....$
$=\frac{1}{2}\Big[1+\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{2}\Big)^6+\ .....\Big]$
$=\frac{1}{2}\Bigg[\frac{1}{1-\Big(\frac{1}{2}\Big)^3}\Bigg] \Big[\text{Since S}_\infty =\frac{\text{A}}{1-\text{r}}\text{ for G.P.}\Big]$
$=\frac{1}{2}\bigg[\frac{1}{1-\frac{1}{8}}\bigg]$
$=\frac{1}{2}\Big[\frac{8}{7}\Big]$
$=\frac{4}{7}$
B wins in $2^{nd}, 5^{th}, 8^{th},$ ..... toss of coin
P(B wins)
$=\text{P}(\overline{\text{E}}\cap\text{E}\cup\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cup\text{E}\ .....)$
$=\text{P}(\overline{\text{E}}\cap\text{E})+\text{P}(\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cap\overline{\text{E}}\cup\text{E}\cup)+\ .....$
$=\text{P}(\overline{\text{E}})\text{P}(\text{E})+\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\text{E})+\ .....$
$=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\ .....$
$=\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^5+\ .....$
$=\Big(\frac{1}{2}\Big)^2\Big[1+\Big(\frac{1}{2}\Big)^3+\ .....\Big]$
$=\frac{1}{4}\Bigg[\frac{1}{1+\Big(\frac{1}{2}\Big)^3}\Bigg]$
$=\frac{1}{4}\bigg[\frac{1}{1-\frac{1}{8}}\bigg]$
$=\frac{1}{4}\Big[\frac{8}{7}\Big]$
$=\frac{2}{7}$
P(C wins) = 1 - P(A wins) - P(B wins)
$=1-\frac{4}{7}-\frac{2}{7}$
$=\frac{1}{7}$
Probability of winning A, B and C are $\frac{4}{7},\frac{2}{7}$ and $\frac{1}{7}$ respectively.
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Question 55 Marks
A and B take turns in throwing two dice, the first to throw 9 being awarded the prize. Show that their chance of winning are in the ratio 9 : 8.
Answer
Sum of 9 can be obtainde by
E = {(3, 6), (4, 5), (5, 4), (6, 3)}
Probability of throwing $9=\frac{4}{36}$
$\text{P(E)}=\frac{1}{9},\text{P}(\overline{\text{E}})=\frac{8}{9}$
$\Rightarrow\ \text{P(A)}=\text{P(B)}=\frac{1}{9}$
$\Rightarrow\ \text{P}(\overline{\text{A}})=\text{P}(\overline{\text{B}})=\frac{8}{9}$
A and B take turns in throwing two dice
Let A starts the game.
P (A wins the game)
$=\text{P}(\text{A}\cup\overline{\text{A}}\cap\overline{\text{B}}\cup\text{A}\cap\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{A}}\cap\overline{\text{B}}\cap\text{A}\cup\ .....)$
$=\text{P}(\text{A})+\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cup\text{A})+\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{A}}\cap\overline{\text{B}}\cap\text{A})+\ .....$
$=\text{P}(\text{A})+\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})\text{P}(\text{A})+\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})\text{P}(\text{A})+\ .....$
$=\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$
$=\frac{1}{9}\Big[1+\Big(\frac{8}{9}\Big)^2+\Big(\frac{8}{9}\Big)^2+\ .....\Big]$
$=\frac{1}{9}\Bigg[\frac{1}{1-\Big(\frac{8}{9}\Big)^2}\Bigg]\ \begin{bmatrix} \text{Since for a G.P. with first term 9 and common ratio r,}\\ \text{S}_{\infty}=\frac{\text{a}}{1-\text{r}} \end{bmatrix}$
$=\frac{1}{9}\bigg[\frac{1}{1-\frac{64}{81}}\bigg]$
$=\frac{1}{9}\Big[\frac{81}{81-64}\Big]$
$=\frac{9}{17}$
P (B wins the game) = 1 - P(A wins the game)
$=1-\frac{9}{17}$
$=\frac{9}{17}$
Chances of winning of A : B
$=\frac{9}{17}:\frac{8}{17}$
$=9:8$
Chances of winning A : B = 9 : 8
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Question 65 Marks
In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.
Answer
Probability of getting six in any toss of a dice $=\frac{1}{6}$
Probability of not getting six in any toss of a dice $=\frac{5}{6}$
A and B toss the die alternatively.
Hence probability of A,s win
$\text{P(A)}+\text{P}(\overline{\text{AB}}\text{A})+\text{P}(\overline{\text{AB}}\ \overline{\text{AB}}\text{A})+\text{P}(\overline{\text{AB}}\ \overline{\text{AB}}\ \overline{\text{AB}}\ \text{A})+\ ......$
$=\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6} \\+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\ .....$
$=\frac{1}{6}+\Big(\frac{5}{6}\Big)^2\frac{1}{6}+\Big(\frac{5}{6}\Big)^4\frac{1}{6}+\Big(\frac{5}{6}\Big)^6\frac{1}{6}+\ .....$
$=\frac{\frac{1}{6}}{1-\Big(\frac{5}{6}\Big)^2}=\frac{1}{6}\times\frac{36}{11}=\frac{6}{11}$
Similarly, probability of B's win
$=\text{P}(\overline{\text{A}}\text{B})+\text{P}(\overline{\text{AB}}\ \overline{\text{A}}\ \text{B})+\text{P}(\overline{\text{AB}}\ \overline{\text{AB}}\ \overline{\text{A}}\ \text{B}) + \ .....$
$=\frac{5}{6}\times\frac{1}{6}+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6} \\+\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}+\ .....$
$=\frac{5}{6}\times\frac{1}{6}+\Big(\frac{5}{6}\Big)^2\frac{5}{6}\times\frac{1}{6}+\Big(\frac{5}{6}\Big)^4\times\frac{5}{6}\times\frac{1}{6}\\+\Big(\frac{5}{6}\Big)^6\times\frac{5}{6}\times\frac{1}{6}+\ .....$
$=\frac{\frac{5}{6}\times\frac{1}{6}}{1-\Big(\frac{5}{6}\Big)^2}=\frac{5}{36}\times\frac{36}{11}=\frac{5}{11}$
Since the probabilities are not equal,
The decision of the refree was not a fair one.
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Question 75 Marks
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts the game, find the probability that B will win the game.
Answer
Let E be event of occuring head in a toss of fair coin.
$\text{P(E)}=\frac{1}{2}$
$\text{P}(\overline{\text{E}})=\frac{1}{2}$
A wins the game in first of $3^{rd}$ or $5^{th}$ throw, .....
Probability that A wins in first throw
$=\text{P(E)}=\frac{1}{2}$
Probability that A wins in $2^{rd}$ throw
$=\text{P}(\overline{\text{E}})\text{ P}(\overline{\text{E}})\text{ P(E)}$
$=\Big(\frac{1}{2}\Big)^2\Big(\frac{1}{2}\Big)$
$=\Big(\frac{1}{2}\Big)^3$
Probability that A wins is $5^{th}$ throw
$=\text{P}(\overline{\text{E}})\text{ P}(\overline{\text{E}})\text{ P}(\overline{\text{E}})\text{ P}(\overline{\text{E}})\text{ P(E)}$
$=\Big(\frac{1}{2}\Big)^4\Big(\frac{1}{2}\Big)$
$=\Big(\frac{1}{2}\Big)^5$
Hence,
Probability of winning A
$=\frac{1}{2}+\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{2}\Big)^5+\ .....$
$=\frac{1}{2}\Big[1+\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^4+\ .....\Big]$
$=\frac{1}{2}\bigg[\frac{1}{1-\big(\frac{1}{2}\big)^2}\bigg]\ \Big[\text{Since S}_\infty=\frac{\text{a}}{1-\text{r}}\text{for G.P.}\Big]$
$=\frac{1}{2}\bigg[\frac{1}{1-\frac{1}{4}}\bigg]$
$=\frac{1}{2}\times\frac{4}{3}$
$=\frac{2}{3}$
Probability that B wins = 1 - P (A wins)
$=1-\frac{2}{3}$
$=\frac{1}{3}$
Required probability $=\frac{1}{3}$
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Question 85 Marks
Three persons $A, B, C$ throw a die in succession till one gets a 'six' and wins the game. Find their respective probabilities of winning.
Answer
Let E be the even of getting a six
$\text{P(E)}=\frac{1}{6}$
$\text{P}(\overline{\text{E}})=\frac{5}{6}$
A wins if he gets a six in $1^{\text {st }}$ or $4^{\text {th }}, 7^{\text {th }} . . .$. throw
A wins in first throw $\mathrm{P}(\mathrm{E})=\frac{1}{6}$
A wins is $4^{\text {th }}$ throw if he fails in $1^{\text {st }}, \mathrm{B}$ fails in $2^{\text {nd }}, \mathrm{C}$ fails in $3^{\text {rd }}$ throw.
Probability of winning A in $4^{\text {th }}$ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{5}{6}\Big)^3\times\frac{1}{6}$
Hence, probability of winning of A
$=\frac{1}{6}+\Big(\frac{5}{6}\Big)^3\times\frac{1}{6}+\Big(\frac{5}{6}\Big)^6\times\frac{1}{6}+\ .....$
$=\frac{1}{6}\Big[1+\Big(\frac{5}{6}\Big)^3+\Big(\frac{5}{6}\Big)^6+\ .....\Big]$
$=\frac{1}{6}\bigg[\frac{1}{1-\big(\frac{3}{5}\big)^3}\bigg]\Big[\text{Using S}_{\infty}=\frac{\text{a}}{1-\text{r}}\ \text{for G. P.}\Big]$
$=\frac{1}{6}\bigg[\frac{1}{1-\frac{125}{216}}\bigg]$
$=\frac{1}{6}\times\frac{216}{91}$
$=\frac{36}{91}$
B wins if he gets a six in $2^{\text {nd }}$ or $5^{\text {th }}$ or $8^{\text {th }}$ $\qquad$ throw.
B wins in $2^{\text {nd }}$ throw $=\mathrm{P}(\overline{\mathrm{E}}) \mathrm{P}(\mathrm{E})$
$=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)$
B wins in $5^{\text {th }}$ throw if a fails in first, $B$ fails in $2^{\text {nd }}, C$ fails in $3^{\text {rd }}, A$ fails in $4^{\text {th }}$.
Probabiliy of winning $B$ in $5^{\text {th }}$ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{5}{4}\Big)^4\Big(\frac{1}{6}\Big)$
Probability of winning B in $8^{th}$ thrwo
$=\Big(\frac{5}{6}\Big)^7\Big(\frac{1}{6}\Big)$
Hence, probability of winning B
$=\Big(\frac{5}{6}\Big)\frac{1}{6}+\Big(\frac{5}{6}\Big)^4\Big(\frac{1}{6}\Big)+\Big(\frac{5}{6}\Big)^7\Big(\frac{1}{6}\Big)$
$=\frac{5}{6}\times\frac{1}{6}\Big[1+\Big(\frac{5}{6}\Big)^3+\Big(\frac{5}{6}\Big)^6+\ .....\Big]$
$=\frac{5}{36}\bigg[\frac{1}{1-\big(\frac{5}{6}\big)^3}\bigg]\Big[\text{Since S}_{\infty}=\frac{\text{a}}{1-\text{r}}\ \text{for G.P.}\Big]$
$=\frac{5}{36}\bigg[\frac{1}{1-\frac{125}{216}}\bigg]$
$=\frac{5}{36}\times\Big[\frac{216}{91}\Big]$
$=\frac{30}{91}$
Probability of winning C = 1 - P(A wins) - P(B wins)
$=1-\frac{36}{91}-\frac{30}{91}$
The probabilities of winning of A, B, and C are $\frac{36}{91},\frac{30}{91}$ and $\frac{25}{91}$.
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Question 95 Marks
An insurance company insured $3000$ scooters, $4000$ cars and $5000$ trucks. The probabilities of the accident involving a scooter, a car and a truck are $0.02, 0.03$ and $0.04$ respectively. One of the insured vehicles meet with an accident. Find the probability that it is a,
  1. Scooter.
  2. Car.
  3. Truck.
Answer
Let $E_1, E_2$ and $E_3$ denote the events that the vehicle is a scooter, a car and a truck, respectively.
Let A be the event that the vehicle meets with an accident.
It is given that there are 3000 scooters, 4000 cars and 5000 trucks.
Total number of vehicles = 3000 + 4000 + 5000 = 12000
$\text{P}(\text{E}_1)=\frac{3000}{12000}=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{4000}{12000}=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{5000}{12000}=\frac{5}{12}$
The probability that the vehicle, which meets with an accident, is a scooter is given by $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big).$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=0.02=\frac{2}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=0.03=\frac{3}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=0.04=\frac{4}{100}$
Using Baues's theorem, we get
  1. Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{4}\times\frac{2}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{1}{2}}{\frac{1}{2}+1+\frac{5}{3}}=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}=\frac{3}{19}$
  1. Required probability $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\times\frac{3}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{1}{2}}{\frac{1}{2+1+\frac{5}{3}}}=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}=\frac{6}{19}$
  1. Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{5}{15}\times\frac{4}{100}}{\frac{1}{4}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{5}{12}\times\frac{4}{100}}$

$=\frac{\frac{5}{3}}{\frac{1}{2}+1+1\frac{5}{3}}=\frac{\frac{5}{3}}{\frac{3+6+10}{6}}=\frac{10}{19}$
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Question 105 Marks
The contents of urns I, II, III are as follows:
Urn I : 1 white, 2 black and 3 red balls
Urn II : 2 white, 1 black and 1 red balls
Urn III : 4 white, 5 black and 3 red balls.
One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I, II, III?
Answer
Urn I contains 1 white, 2 black and 3 red balls
Urn II contains 2 white, 1 black and 1 red balls
Urn III contains 4 white, 5 black and 3 red balls.
Consider $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ and A be events as:
$\mathrm{E}_1=$ Selecting unr I
$\mathrm{E}_2=$ Selecting urn II
$\mathrm{E}_3=$ Selecting urn III
$\mathrm{A}=$ Drawing 1 white and 1 red balls
$\text{P}(\text{E}_1)=\frac{1}{3}$
$\text{P}(\text{E}_2)=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{1}{3}$
$P (A|E_1)$ = P[Drawing 1 red and 1 white from urn I]
$=\frac{^{1}\text{C}_1\times ^{3}\text{C}_1}{^{6}\text{C}_2}$
$=\frac{1\times3}{\frac{6\times5}{2}}$
$=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing 1 red and 1 white from urn II]
$=\frac{^{2}\text{C}_1\times ^{1}\text{C}_1}{^{4}\text{C}_2}$
$=\frac{2\times1}{\frac{4\times3}{2}}$
$=\frac{1}{3}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ [Drawing 1 red and 1 white from urn III]
$=\frac{^{2}\text{C}_1\times ^{1}\text{C}_1}{^{4}\text{C}_1}$
$=\frac{2\times1}{\frac{12\times11}{2}}$
$=\frac{2}{11}$
We have to find,
P(They come from urn I) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
P(They come from urm II) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)$
P(They come from urn III) $=\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)$
By baye,s theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\times\frac{1}{5}}{\frac{1}{3}\times\frac{1}{5}+\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{2}{11}}$
$=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}$
$=\frac{\frac{1}{5}}{\frac{36+55+30}{165}}$
$=\frac{1}{5}\times\frac{165}{118}$
$=\frac{33}{118}$
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\times\frac{1}{3}}{\frac{1}{3}\times\frac{1}{5}+\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{2}{11}}$
$=\frac{\frac{1}{3}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}$
$=\frac{\frac{1}{3}}{\frac{33+55+30}{165}}$
$=\frac{1}{3}\times\frac{165}{118}$
$=\frac{55}{118}$
$\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\times\frac{2}{11}}{\frac{1}{3}\times\frac{1}{5}+\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{2}{11}}$
$=\frac{\frac{2}{11}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}$
$=\frac{2}{11}\times\frac{165}{118}$
$=\frac{30}{118}$
Therefore, required probability $=\frac{33}{118},\frac{55}{118},\frac{30}{118}.$
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Question 115 Marks
A manufacturer has three machine operators $A, B$ and $C$. The first operator A produces $1 \%$ defective items, whereas the other two operators B and C produce $5 \%$ and $7 \%$ defective items respectively. A is on the job for $50 \%$ of the time, $B$ on the job for $30 \%$ of the time and C on the job for $20 \%$ of the time. A defective item is produced. What is the probability that it was produced by $A$?
Answer
Let $E_1, E_2,$ and $E_3$ be the respective events of the time consumed by machine A, B, and C for the job.
$\text{P}(\text{E}_1)=50\%=\frac{50}{100}=\frac{1}{2}$
$\text{P}(\text{E}_2)=30\%=\frac{30}{100}=\frac{3}{10}$
$\text{P}(\text{E}_3)=20\%=\frac{20}{100}=\frac{1}{5}$
Let X be the event of producing defective items.
$\text{P}(\text{X}|\text{E}_1)=1\%=\frac{1}{100}$
$\text{P}(\text{X}|\text{E}_2)=5\%=\frac{5}{100}$
$\text{P}(\text{X}|\text{E}_3)=7\%=\frac{7}{100}$
The probability that the defective item was produced by A is given by $P(E_1|A)$.
By using Bayes' theorem, we obtain
$\text{P}(\text{E}_1|\text{X})=\frac{\text{P}\text{E}_1\text{P}(\text{X}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{X}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{X}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{X}|\text{E}_3)}$
$=\frac{\frac{1}{2}\times\frac{1}{100}}{\frac{1}{2}\times\frac{1}{100}+\frac{3}{10}\times\frac{5}{100}+\frac{1}{5}\times\frac{7}{100}}$
$=\frac{\frac{1}{100}\times\frac{1}{2}}{\frac{1}{100}\Big(\frac{1}{2}+\frac{3}{2}+\frac{7}{5}\Big)}$
$=\frac{\frac{1}{2}}{\frac{17}{5}}$
$=\frac{5}{34}$
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Question 125 Marks
The contents of three urns are as follows:
Urn 1 : $7$ white, $3$ black balls,
Urn 2 : $4$ white, $6$ black balls,
Urn 3 : $2$ white, $8$ black balls.
One of these urns is chosen at random with probabilities $0.20, 0.60$ and $0.20$ respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn $3$?
Answer
Urn I contains $7$ white and $3$ black balls
Urn II contains $4$ white and $6$ black balls
Urn III contains $2$ white and $8$ black balls
Let $E_1, E_2, E_3$ and A be evets as:
$E_1$​​​​​​​ = Selecting urn I
$E_2​​​​​​​$ = Selecting urn II
$E_3​​​​​​​$​​​​​​​ = Selecting urn III
A = Drawing $2$ white balls without replacement.
Given,
$P(E_1) = 0.20$
$P(E_2) = 0.60$
$P(E_3) = 0.20$
$P(A|E_1)$ = P[Drawing $2$ white ball from urn $I$]
$=\frac{^{7}\text{C}_2}{^{10}\text{C}_2}$
$=\frac{\frac{7\times6}{2}}{\frac{10\times9}{2}}$
$=\frac{7}{15}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing $2$ white ball from urn $II$]
$=\frac{^{4}\text{C}_2}{^{10}\text{C}_2}$
$=\frac{\frac{4\times3}{2}}{\frac{10\times9}{2}}$
$=\frac{12}{90}$
$=\frac{2}{15}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ [Drawing $2$ white ball from urn $III$]
$=\frac{^{2}\text{C}_2}{^{10}\text{C}_2}$
$=\frac{1}{\frac{10\times9}{2}}$
$=\frac{1}{45}$
To find
P(2 white balls drawn are from urn III) $=\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{0.2\times\frac{1}{45}}{0.2\times\frac{7}{15}+0.6\times\frac{2}{15}+0.2\times\frac{1}{45}}$
$=\frac{\frac{2}{450}}{\frac{14}{150}+\frac{12}{150}+\frac{2}{450}}$
$=\frac{\frac{2}{450}}{\frac{42+36+2}{450}}$
$=\frac{2}{80}$
$=\frac{1}{40}$
Required probability $=\frac{1}{40}$
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Question 135 Marks
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the constitutional probability that both are girls? Given that.
  1. The youngest is a girl,
  2. At least one is a girl.
Answer
  1. Let 'A' be the event that both the children born are girls. Let 'B' be the event that the youngest is a girls. We have to find conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$

$\text{A}\subset\text{B} \Rightarrow\text{A}\cap\text{B}=\text{A}$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

$\text{P(B)}=\text{P(BG)}+\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2} \\=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
  1. Let 'A' be the event that both the children born are girls. Let 'B' be the event that at least one is a girl. We have to find the conditional probability $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$

$\text{A}\subset\text{B}\Rightarrow\text{A}\cap\text{B}=\text{A}$

$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}=\text{P(GG)}=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

$\text{P(B)}=1-\text{P(BB)}=1-\frac{1}{2}\times\frac{1}{2}=1-\frac{1}{4}=\frac{3}{4}$

Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
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Question 145 Marks
Three urns contains $2$ white and $3$ black balls; $3$ white and $2$ black balls and $4$ white and $1$ black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn.
Answer
Urn I contains 2 white and 3 black balls
Urn II contains 3 white and 2 black balls
Urn III contains 4 white and 1 black balls.
Let $E_1, E_2, E_3$ and $A$ be events as:
$\mathrm{E}_1=$ Selecting urn I
$\mathrm{E}_2=$ Selecting urn II
$\mathrm{E}_3=$ Selecting urn III
A = A white balls is drawn
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{E}_3)=\frac{1}{2}$ [Since there are 3 urns]
$P (A | E_1)$ = P [Drawing 1 white ball from uen 1]
$=\frac{2}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing 1 white ball from urn II]
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ [Drawing one white ball from urn III]
$=\frac{4}{5}$
To find,
P (Drawn one white ball from urm I) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{4}{5}}$
$=\frac{\frac{2}{10}}{\frac{2+3+4}{10}}$
$=\frac{2}{9}$
Required probability $=\frac{2}{9}$
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Question 155 Marks
A coin is tossed three times. Let the events A, B and C be defined as follows:
A = first toss is head, B = second toss is head, and C = exactly two heads are tossed in a row. Check the independence of,
  1. A and B.
  2. B and C.
  3. C and A.
Answer
A coin is tossed three times,
Sample space = {HHH, HTH, THH, HHT, HTT, THT, TTT}
A = First toss is head
A = {HHH, HHT, HTH, HTT}
$\text{P(A)}=\frac{4}{8}$
$\text{P(A)}=\frac{1}{2}$
B = Second toss is head
= {HHH, HHT, THH, THT}
$\text{P(B)}=\frac{4}{8}$
$\text{P(B)}=\frac{1}{2}$
C = exactly two head in a row
C = {HHT, THH}
$\text{P(C)}=\frac{2}{8}$
$\text{P(C)}=\frac{1}{4}$
$\text{A}\cap\text{B}\{\text{HHH, HHT}\}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}$
$=\frac{1}{4}$
$\text{B}\cap\text{C}=\{\text{HHT, THH}\}$
$\text{P}(\text{B}\cap\text{C})=\frac{2}{8}$
$\text{P}(\text{B}\cap\text{C})=\frac{1}{4}$
$(\text{A}\cap\text{C})=\{\text{HHT}\}$
$\text{P}(\text{A}\cap\text{C})=\frac{1}{8}$
  1. $\text{P(A)} \text{ P(B)}=\frac{1}{2}\times\frac{1}{2}$
$=\frac{1}{4}$

$\text{P(A) }\text{P(B)}=\text{P}(\text{A}\cap\text{B})$

Hence, A and B are independent events.
  1. $\text{P(B) }\text{P(C)}=\frac{1}{2}\times\frac{1}{4}$
$=\frac{1}{8}$

$\text{P(B) }\text{P(C)}\neq\text{P}(\text{B}\cap\text{C})$

So, B and C are not independent events.
  1. $\text{P(A) }\text{P(C)}=\frac{1}{2}\times\frac{1}{4}$
$=\frac{1}{8}$

$\text{P(A) }\text{P(C)}=\text{P}(\text{A}\cap\text{C})$

Hence, A and C are independent events.
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Question 165 Marks
A bag $A$ contains $2$ white and $3$ red balls and a bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag $B.$
Answer
Bag A contains 2 white and 2 red balls Bag B
contains 4 white and 5 red balls.
Consider $E _1, E _2$ and A events as:
$E_1 =$ Selecting bag $A$
$E_2 =$ Selecting bag $B$
A = Drawing one red ball
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since there are 2 bags]
$P (A|E_1) = P $[Drawing one red ball from bag A]
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing one re ball from bag B]
$=\frac{5}{9}$
To find,
P (Drawn, one red ball is from bag B) $=\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
By baye's theorem
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{5}{9}}{\frac{1}{2}\times\frac{3}{5}\times\frac{1}{2}\times\frac{5}{9}}$
$=\frac{\frac{5}{9}}{\frac{3}{5}+\frac{5}{9}}$
$=\frac{\frac{5}{9}}{\frac{3}{5}+\frac{5}{9}}$
$=\frac{\frac{5}{9}}{\frac{27+25}{45}}$
$=\frac{5}{9}\times\frac{45}{52}=\frac{25}{52}$
Required probability $=\frac{25}{52}$
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Question 175 Marks
A factory has three machines $\mathrm{X}, \mathrm{Y}$ and Z producing 1000,2000 and 3000 bolts per day respectively. The machine X produces $1 \%$ defective bolts, $Y$ produces $1.5 \%$ and $Z$ produces $2 \%$ defective bolts. At the end of a day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?
Answer
Consider $E_1, E_2$ and A events as:
$E_1$ = Bolt produced by machine X
$E_2$ = Bolt produced by machine Y
$E_3$ = Bolt produced by machine Z
A = A bolt drawn is defective.
$\text{P}(\text{E}_1)=\frac{1000}{6000}=\frac{1}{6}$
$\text{P}(\text{E}_2)=\frac{2000}{6000}=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{1000}{3000}=\frac{1}{2}$
$P(A|E_1)$ = P(Drawing defective bolt from machine Y)
$=\frac{1}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Drawing defective bolt from machine Y)
$=\frac{0.5}{100}$
$=\frac{3}{200}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Drawing defective bolt from machine Z)
$=\frac{2}{100}$
To find, P(Defective bolt frawn is produced by machine X) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
Bu baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{6}\times\frac{1}{100}}{\frac{1}{6}\times\frac{1}{100}+\frac{1}{3}\times\frac{3}{200}+\frac{1}{2}\times\frac{2}{100}}$
$=\frac{\frac{1}{600}}{\frac{1}{600}+\frac{3}{600}+\frac{1}{100}}$
$=\frac{1}{10}$
Required probability $=\frac{1}{10}$
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Question 185 Marks
In a class, $5 \%$ of the boys and $10 \%$ of the girls have an IQ of more than $150$. In this class, $60 \%$ of the students are boys. If a student is selected at random and found to have an IQ of more than $150$, find the probability that the student is a boy.
Answer
Consider $E_1, E_2$ and A events as:
$E_1$ = Selected students is boy
$E_2$ = Selected Students is girl
$E_3$ = A students with IQ more that 150 is selected
$\text{P}(\text{E}_1)=\frac{60}{100}$
$\text{P}(\text{E}_2)=\frac{40}{100}$
$\text{P}(\text{A}|\text{E}_1)=\text{P}$
(Selected Boy has IQ more than 150)
$=\frac{5}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selected girl has IQ more than 150)
$=\frac{10}{100}$
To find, P (Selected student with IQ more than 150 is a boy) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big).$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{60}{100}\times\frac{5}{100}}{\frac{60}{100}\times\frac{5}{100}+\frac{40}{100}\times\frac{10}{100}}$
$=\frac{300}{300+400}$
$=\frac{300}{700}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
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Question 195 Marks
A bag A contains $5$ white and $6$ black balls. Another bag B contains $4$ white and $3$ black balls. A ball is transferred from bag A to the bag B and then a ball is taken out of the second bag. Find the probability of this ball being black.
Answer
Given,
Bag A contains $5$ white and $6$ black balls.
Bag B contains $4$ white and $3$ black balls.
There are two ways of transferring a ball from bag A to bag B
I - By transferring one white ball from A ot bag B then drawing one black ball from bag B.
II - By transferring one black ball from bag A to bag B, then drawing one black from bag B.
Let, $E_1, E_2$​​​​​​​ and A be events as below:
$E_1​​​​​​​$​​​​​​​ = One black ball drawn from bag A
$E_2​​​​​​​$​​​​​​​ = One black ball drawn from bag B
A = One black ball drawn from bag B
$\text{P}(\text{E}_1)=\frac{5}{11}$
$\text{P}(\text{E}_2)=\frac{6}{11}$
$=\text{P}(\text{A}|\text{E}_1)=\frac{3}{8}$
[Since, $E_1​​​​​​​$​​​​​​​ has increased one white ball in bag B]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{4}{8}$
[Since, $E_2​​​​​​​$​​​​​​​ has increased one black ball in bag B]
By the law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{5}{11}\times\frac{3}{8}+\frac{6}{11}\times\frac{4}{8}$
$=\frac{15}{88}+\frac{24}{88}$
$=\frac{39}{88}$
Required probability $=\frac{39}{88}$
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Question 205 Marks
A factory has three machines $A, B$ and $C$, which produce $100, 200$ and $300$ items of a particular type daily. The machines produce $2 \%, 3 \%$ and $5 \%$ defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A .
Answer
Let $E_1, E_2, E_3$ and A be events as:
$E_1$ = Selecting product from machine A
$E_2$ = Selecting product from machine B
$E_3$ = Selecting product from machine C
A = Selecting a defective product
$\text{P}(\text{E}_1)=\frac{100}{600}=\frac{1}{6}$
$\text{P}(\text{E}_2)=\frac{200}{600}=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{300}{600}=\frac{1}{2}$
$P(A|E_1)$ = P(Selecting a defective item from machine A)
$=\frac{2}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting a defective item from machine B)
$=\frac{3}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting a defective item machine C)
$=\frac{5}{100}$
To find, P(Selecting defective item is produced by machine A) $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{6}\times\frac{2}{100}}{\frac{1}{6}\times\frac{2}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{1}{2}\times\frac{5}{100}}$
$=\frac{\frac{2}{600}}{\frac{2}{600}+\frac{3}{600}+\frac{5}{200}}$
$=\frac{2}{600}\times\frac{600}{23}$
$=\frac{2}{23}$
Required probability $=\frac{2}{23}$
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Question 215 Marks
Three urns $A, B$ and $C$ contain $6$ red and $4$ white; $2$ red and $6$ white; and $1$ red and $5$ white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A.
Answer
Urn A conrains 6 red and 4 white balls
Urn B contains 2 red and 6 white balls
Urn C contains 1 red and 5 white balls
Consider $E_1, E_2, E_3$ and A events as:
$E_1$ = Selecting urn A
$E_2$ = Selecting urn B
$E_3$ = Selecting urn C
A = Selecting ared ball
$\text{P}(\text{E}_1)=\frac{1}{3}$
$\text{P}(\text{E}_2)=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{1}{3}$ [Since there are three urns]
$P(A|E_1)$ = P(Selecting a red ball from urn A)
$=\frac{6}{10}$
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting a red ball from urn B)
$=\frac{2}{8}$
$=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Selecting a red ball from urn C)
$=\frac{1}{6}$
To find, P(Selected red ball is from urn A) $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\times\frac{3}{5}}{\frac{1}{3}\times\frac{3}{5}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{3}\times\frac{1}{6}}$
$=\frac{\frac{3}{4}}{\frac{3}{5}+\frac{1}{4}+\frac{1}{6}}=\frac{36}{61}$
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Question 225 Marks
An insurance company insured $2000$ scooters and $3000$ motorcycles. The probability of an accident involving a scooter is $0.01$ and that of a motorcy is $0.02$. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
Answer
Let $E_1, E_2$ and A be events ar:
$E_1$ = Vehilcle is scooter
$E_2$ = Vehicle is motorcycle
A = An insured met with scooter
$\text{P}(\text{E}_1)=\frac{2000}{5000}=\frac{2}{5}$
$\text{P}(\text{E}_2)=\frac{3000}{5000}=\frac{3}{5}$
$P(A|E_1)$ = P(Accident of scooter)
= 0.01
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ P(Accident of motorcycle)
= 0.02
To find, P(Accident vehicle was motorcycle) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{5}\times\frac{2}{100}}{\frac{2}{5}\times\frac{1}{100}+\frac{3}{5}\times\frac{2}{100}}$
$=\frac{\frac{6}{500}}{\frac{2}{500}+\frac{6}{500}}$
$=\frac{6}{8}$
$=\frac{3}{4}$
Required probability $=\frac{3}{4}$
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Question 235 Marks
The contents of three bags I, II and III are as follows:
Bag I : $1$ white, $2$ black and $3$ red balls,
Bag II : $2$ white, $1$ black and $1$ red ball;
Bag III : $4$ white, $5$ black and $3$ red balls.
A bag is chosen at random and two balls are drawn. What is the probability that the balls are white and red?
Answer
A white ball and a red ball can be drawn in three mutually exclusice ways:
  1. Selecting bag I and then drawing a white and a red ball from it.
  2. Selecting bag II and then drawing a white and a red ball from it.
  3. Selecting bag III and then drawing a white and a red ball from it.
Let $E_1, E_2$​​​​​​​ and A be the events as defined below;
$E_1​​​​​​​$​​​​​​​ = Selecting bag I
$E_2​​​​​​​$​​​​​​​ = Selecting bag II
$E_3​​​​​​​$​​​​​​​ = Selecting bag III
A = Drawing A white and a red ball
It is given that one of the bags is selected randomly.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{3}$
$\text{P}(\text{E}_2)=\frac{1}{3}$
$\text{P}(\text{E}_3)=\frac{1}{3}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{^{1}\text{C}_1\times ^{3}\text{C}_1}{^{6}\text{C}_2}=\frac{3}{15}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{^{2}\text{C}_1\times ^{1}\text{C}_1}{^{4}\text{C}_2}=\frac{2}{6}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{^{4}\text{C}_1\times ^{3}\text{C}_1}{^{12}\text{C}_2}=\frac{12}{66}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E})_1\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)$
$=\frac{1}{3}\times\frac{3}{15}+\frac{1}{3}\times\frac{2}{6}+\frac{1}{3}\times\frac{12}{66}$
$=\frac{1}{15}+\frac{1}{9}+\frac{2}{33}$
$=\frac{33+55+30}{495}$
$=\frac{118}{495}$
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Question 245 Marks
Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and third card drawn is an ace?
Answer
Let K denote the event that the card drawn is king and a be the event taht the card drawn is an ace.
We are ot find P (K K A).
Now, $\text{P(K)}=\frac{4}{52}$
Also, $\text{P}\Big(\frac{\text{K}}{\text{K}}\Big)$ is the probability of second king with the condition that one king has already been drawn.
Now, there are 3 king in (52 - 1) = 51 cards.
$\therefore\ \text{P} \Big(\frac{\text{K}}{\text{K}}\Big)=\frac{3}{51}$
Lastly, $\text{P}\Big(\frac{\text{A}}{\text{K K}}\Big)$ is the probability of third drawn card to be an ace, woth the condition that two kings have already been drawn.
Now, there are four aces in left 50 cards.
$\therefore\ \text{P}\Big(\frac{\text{A}}{\text{K K}}\Big)=\frac{4}{50}$
By multiplication law of probability, we have
$\text{P(K K A)}=\text{P(K) P}\Big(\frac{\text{K}}{\text{K}}\Big) \text{ P}\Big(\frac{\text{A}}{\text{KK}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}\times\frac{4}{50}=\frac{2}{5525}$
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Question 255 Marks
An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?
Answer
Given,
An anti aircraft gun can rake a maximum 4 shots at an enemy plane
Consider,
A = Htting the plane at first shot
B = Hetting the plane at second shot
C = Hetting the place at third shot
D = Hetting the place at fourth shot
⇒ P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, P(D) = 0.1
P (Gun hits the place)
= 1 - P(Gun does not hit the plane)
= 1 - P(Non of the foru shots hot the place)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}}\cap\overline{\text{C}}\cap\overline{\text{D}})$
$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})\text{ P}(\overline{\text{C}})\text{ P}(\overline{\text{D}})$
$=1-[1-\text{P(A)}]\big[1-\text{P}(\overline{\text{B}})\big][1-\text{P(C)}][1-\text{P(D)}]$
$=1-[1-0.4][1-0.3][1-0.2][1-0.1]$
$=1-(0.6)(0.7)(0.8)(0.9)$
$=1.03024$
$=0.6976$
Required probability = 0.6976
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Question 265 Marks
Arun and Tarun appeared for an interview for two vacancies. The probability of Arun's selection is $\frac{1}{4}$ and that to Tarun's rejection is $\frac{2}{3}$. Find the probability that at least one of them will be selected.
Answer
Given,
Probability of Arun's (A) selection $=\frac{1}{4}$
$\text{P(A)}=\frac{1}{4}$
Probability of tarun's (T) rejection $=\frac{2}{3}$
$\text{P}(\overline{\text{T}})=\frac{2}{3}$
$\text{P}(\overline{\text{A}})=1-\text{P(A)}$
$\Rightarrow\ \text{P}(\overline{\text{A}})=1-\frac{1}{4}$
$\Rightarrow\ \text{P}(\overline{\text{A}})=\frac{3}{4}$
$\text{P(T)}=1-\text{P}(\overline{\text{T}})$
$\Rightarrow\ \text{P(T)}=1-\frac{2}{3}$
$\Rightarrow\ \text{P(T)}=\frac{1}{3}$
P (At least one of them will be selelcted)
= 1 - P(None of them selected)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{T}})$
$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{T}})$
$=1-\frac{2}{3}\times\frac{3}{4}$
$=\frac{1}{2}$
Required probability $=\frac{1}{2}$
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Question 275 Marks
One bag contains $4$ white and $5$ black balls. Another bag contains $6$ white and $7$ black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is white.
Answer
There are two bags.
Bag (1) contain 4 white and 5 black balls.
Bag (2) contain 6 white and 7 black balls.
A ball is taken from bag (1) and without seeing its colour is pur in bag (2). Then a ball is drawn from bag (2) and is found white in colour.
P(1 white ball from bag 1) $=\frac{4}{9}$
$\text{P}(\text{W}_1)=\frac{4}{9}$
P(1 black ball from bag 1) $=\frac{5}{9}$
$\text{P}(\text{B}_1)=\frac{5}{9}$
P(1 white ball from bag 2 given $W_1$ is put in bag 2)
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)=\frac{7}{14}$
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)=\frac{1}{2}$
P(1 white ball from bag 2 given $B_1$ is put in bag 2)
$\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)=\frac{6}{14}$
P(1 white from bag 2)
$=\text{P}(\text{W}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)+\text{P}(\text{B}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)$
$=\frac{4}{9}\times\frac{1}{2}+\frac{5}{9}\times\frac{6}{14}$
$=\frac{4}{18}+\frac{30}{126}$
$=\frac{58}{126}$
$=\frac{29}{63}$
Required probability $=\frac{29}{63}$
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Question 285 Marks
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
Answer
Let E and F denote respectively the events that first and second ball drawn are black. We have to find $\text{P}(\text{E}\cap\text{F})$ or P(EF)
Now P(E) = P (black ball in first draw) $=\frac{10}{15}$
Also given that the first ball drawn is black, i.e, event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
i.e., $\text{P}(\text{F}|\text{E})=\frac{9}{14}$
By multiplication rule of probability, we have
$\text{P}(\text{E}\cap\text{F})=\text{P}(\text{E})\text{ P}(\text{F}|\text{E})$
$=\frac{10}{15}\times\frac{9}{14}=\frac{3}{7}$
Multiplication rule of probability for more than two events if E,F and G are three events of sample space, we have
$\text{P}(\text{E}\cap\text{F}\cap\text{G})=\text{P}(\text{E})\text{ P}(\text{F}|\text{E}) \text{ P}(\text{G}) (\text{G}\cap\text{F})= \text{P}(\text{E})\text{ P}(\text{F}|\text{E}) \text{ P}(\text{G}|\text{EF})$
Similarly, the multiplication rule of probability can be extended for four or more events.
The following example illustrates the extension of multiplication rule of probability for three events.
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Question 295 Marks
There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?
Answer
A be the event of choosing two - headed coin,
B be the event of choosing a biased coin that comes up head 75% of the times,
C be the event of choosing a biased coin that comes up tail 40% of the times and
E be the event of getting a head.
Now,
$\text{P(A)}=\text{P(B)}=\text{P(C)}=\frac{1}{3}$ and
$\text{P}(\text{E}|\text{A})=1,\text{P}(\text{E}|\text{B})=75\%=\frac{75}{100}=\frac{3}{4}$ and $\text{P}(\text{E}|\text{C})=60\%=\frac{60}{100}=\frac{3}{5}$
So, using Bayes' theorem, we get
P (the head shown was of two - headed coin) = P(A|E)
$=\frac{\text{P(A)}\times\text{P}(\text{E}|\text{A})}{\text{P(A)}\times\text{P}(\text{E}|\text{A})+\text{P(B)}\times(\text{E}|\text{B})+\text{P(C)}\times\text{P}(\text{E}|\text{C})}$
$=\frac{\Big(\frac{1}{3}\times1\Big)}{\Big(\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{3}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\Big)}$
$=\frac{\Big(\frac{1}{3}\Big)}{\Big(\frac{20+15+12}{60}\Big)}$
$=\frac{\Big(\frac{4}{3}\Big)}{\Big(\frac{47}{60}\Big)}$
$=\frac{60}{3\times47}$
$=\frac{20}{47}$
So, the probability that the head shown was of a two-headed coin is $=\frac{20}{47}$.
Disclaimer: The answer given in the book is incorrect. The same has been corrected here.
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Question 305 Marks
The probabilities of two students A and B coming to the school in time are $\frac{3}{7}$ and $\frac{5}{7}$ respectively. Assuming that the events, 'A coming in time' and 'B coming in time' are independent, find the probability of only one of them coming to the school in time. Write at least one advantage of coming to school in time.
Answer
Given that the events 'A coming in time' and 'B coming in time' are independent.
Let 'A' denote the event of 'A coming in time'.
Then, $'\overline{\text{A}'}$ denotes the complementary event of A.
Similarly define B and $\overline{\text{B}}$.
P(Only one coming in time) $=\text{P}(\text{A}\cap\overline{\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\overline{\text{B}})+\text{P}(\overline{\text{A}})\times\text{P(B)}\ ......$
(Since A and B are independent events)
$=\frac{3}{7}\times\frac{2}{7}+\frac{4}{7}\times\frac{5}{7}=\frac{6}{49}+\frac{20}{49}=\frac{26}{49}$
The advantage of coming to school in time is that you will not miss any part of the lecture and will be able to learn more.
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Question 315 Marks
A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. In which of the following cases are the events A and B independent?
A = The card drawn is a king or queen,
B = the card drawn is a queen or jack.
Answer
A card is drawn from 52 cards
It has 4 kings, 4 queen, 4 jack
A = The card drawn is a king ir a queen
$\text{P(A)}=\frac{4+4}{52}$
$=\frac{8}{52}$
$\text{P(A)}=\frac{2}{13}$
B = the card drawn is a queen or a jack
$\text{P(B)}=\frac{4+4}{52}$
$=\frac{8}{52}$
$=\frac{2}{13}$
$\text{A}\cap\text{B}=$ The card drawn is a queen
$\text{P}(\text{A}\cap\text{B})=\frac{4}{52}$
$=\frac{1}{13}$
$\text{P(A)}\text{ P(B)}=\frac{2}{13}\times\frac{2}{13}$
$=\frac{4}{169}$
$\text{P(A)}\text{ P(B)}\neq\text{P}(\text{A}\cap\text{B})$
Hence, A and B are not independent.
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Question 325 Marks
A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. In which of the following cases are the events A and B independent?
A = the card drawn is black,
B = the card drawn is a king.
Answer
A card is drawn from pack of 52 cards
There are 26 black and four kings in which 2 kings are black.
A = the card drawn is black
$\text{P(A)}=\frac{26}{52}$
$\text{P(A)}=\frac{1}{2}$
B = the card drawn is a king
$\text{P(B)}=\frac{4}{52}$
$=\frac{1}{13}$
$\text{A}\cap\text{B}=$ The card drawn is a black king
$\text{P}(\text{A}\cap\text{B})=\frac{2}{52}=\frac{1}{26}$
$\text{P(A) }\text{P(B)}=\frac{1}{2}\times\frac{1}{13}$
$=\frac{1}{26}$
$\text{P(A) }\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are independent events.
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Question 335 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = The number of heads is odd,
B = The number of tails is odd.
Answer
Sample space for a coin thrown thrice is
= {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = the number of head is odd
A = {HTT, THT, TTH, HHH}
B = the number if tails is odd
B = {THH, HTH, HHT, TTT}
$\text{A}\cap\text{B}=\{\}=\phi$
$\text{P(A)}=\frac{4}{8}=\frac{1}{2}$
$\text{P(B)}=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{0}{8}=0$
$\text{P(A)}.\text{P(B)}=\frac{1}{2}\times\frac{1}{2}$
$=\frac{1}{4}$
$\text{P(A)}.\text{P(B)}\neq\text{P}(\text{A}\cap\text{B})$
So, A and B are not independent events.
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Question 345 Marks
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.
Answer
A = Two numbers on two dice are different
= {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = Sum of numbers on the dice is 4
B = {(1, 3), (2, 2), (3, 1)}
$\text{A}\cap\text{B}=\{(1,3),(3,1)\}$
Required probability $=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}$
$=\frac{2}{30}$
Required probability $=\frac{1}{15}$
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Question 355 Marks
A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.
Answer
Tickets are numbered from 1 to 25
⇒ Total number of tickets = 25
Number of tickets with even numbers on it
= 12 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
A = first ticket with even number
B = second ticket with even number
P (Both tickets will show even number, without replacement)
$=\text{P}(\text{A})\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{12}{25}\times\frac{11}{24}$
$=\frac{11}{50}$
Required probability $=\frac{11}{50}$
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Question 365 Marks
A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from,
LONDON.
Answer
Consider events $E_1, E_2$ and A events As:
$E_1$ = Letters come from LONDON
$E_2$ = Letters come from CLIFTON
$E_3$ = Two consecutive letters visible on the envelope are on
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since letters came either from LONDON or CLIFTON]
$P(A | E_1)$ = P(Two consecutive letters ON from LONDON)
$=\frac{2}{5}$
[Since LONDON has 2 - ON and 5 letters we consider one 'ON' as one letter]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Two consecutive letters On from CLIFTON)
$=\frac{1}{6}$
[Since CLIFTON has one 'ON' nad 6 letters considering ON as one letter]
To find, P (ON visible are from LONDON) $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big).$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}$
$=\frac{\frac{2}{10}}{\frac{2}{10}+\frac{1}{12}}$
$=\frac{2}{10}\times\frac{60}{17}$
$=\frac{12}{17}$
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{12}{17}$
Required probability $=\frac{12}{17}$
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Question 375 Marks
A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.
Answer
There are two bags.
Bag (1) contain 3 red and 5 black balls
Bag (2) contain 6 red and 4 black balls
P (One red ball from bag 1) $=\frac{3}{8}$
$\text{P}(\text{R}_1)=\frac{3}{8}$
P (One black ball from bag 1) $=\frac{5}{8}$
$\text{P}(\text{B}_1)=\frac{5}{8}$
P (One red ball from bag 1) $=\frac{6}{10}$
$\text{P}(\text{R}_2)=\frac{3}{5}$
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Question 385 Marks
If $\text{P}(\text{not B})=0.65, \text{P}(\text{A}\cup\text{B})=0.85$, and A and B are independent events, then find P(A).
Answer
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
Since A, B are independent
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(A)}$
Also, $\text{P(noy B)} = 0.65 \Rightarrow \text{P(B)} = 0.35$
Hence, we have
$0.85 = \text{P(A)} + 0.35 - \text{P(A)} (0.35)$
$\Rightarrow 0.8 = \text{P(A)} [1 - 0.35]$
$\Rightarrow \frac{0.5}{.65}=\text{P(A)}$
$\Rightarrow \text{P(A)} = 0.77$
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Question 395 Marks
The bag $A$ contains $8$ white and $7$ black balls while the bag $B$ contains $5$ white and $4$ black balls. One ball is randomly picked up from the bag $A$ and mixed up with the balls in bag $B$. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.
Answer
Bag A contains 8 white and 7 black balls
Bag B contains 5 white and 4 black balls
Transfer can be done in two ways:
I - A white ball is transferred from bag A to bag B and then onw white ball is drawn from bag B.
II - A black ball is transferred fron bag A to bag, then one white ball is drawn from bag B.
Let $E_1, E_2$ and A be events as:
$E_1$ = One white ball from bag A
$E_2$ = one black ball from bag A
A = One white ball from bag B
$\text{P}(\text{E}_1)=\frac{8}{15}$
$\text{P}(\text{E}_2)=\frac{7}{15}$
$\text{P}(\text{A}|\text{E}_1)=\frac{6}{10}$
[Since $E_1$ has increased white balls in bag B]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{5}{10}$
[Since $E_2$ has increased black ball in bag B]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{8}{15}\times\frac{6}{10}+\frac{7}{15}\times\frac{5}{10}$
$=\frac{48}{150}+\frac{35}{150}$
$=\frac{83}{150}$
Required probability $=\frac{83}{150}$
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Question 405 Marks
Given the probability that A can solve a problem is $\frac{2}{3}$ and the probability that B can solve the same problem is $\frac{3}{5}$. Find the probability that none of the two will be able to solve the problem.
Answer
Given,
Probability that A can solve a problem $=\frac{2}{3}$
$\Rightarrow\ \text{P(A)}=\frac{2}{3}$
$=\text{P}(\overline{\text{A}})=1-\frac{2}{3}$
$\text{P}(\overline{\text{A}})=\frac{1}{3}$
Probability that B can solve the same problem $=\frac{3}{5}$
$\Rightarrow\ \text{P(B)}=\frac{3}{5}$
$\Rightarrow\ \text{P}(\overline{\text{B}})=1-\frac{3}{5}$
$\text{P}(\overline{\text{B}})=\frac{2}{5}$
P(None of them solve the problem)
$=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
$=\frac{1}{2}\times\frac{2}{5}$
$=\frac{2}{15}$
Required probability $=\frac{2}{15}$
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Question 415 Marks
A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Answer
A bag contains 7 white, 5 black and 4 red balls.
Four balls are drawn without replacement
P (At least three balls are black)
= P(3 black balls and one not black or 4 black balls)
= P(3 black and one not black) + P(4 black balls)
$=\frac{{^5}\text{C}_3\times{^{11}}\text{C}_1}{{^{16}}\text{C}_4}+\frac{{^{5}}\text{C}_4}{^{16}\text{C}_4}$
$=\frac{\frac{5!}{3!2!}\times11+\frac{5!}{4!1!}}{\frac{16!}{4!12!}}\ \Big[\text{Since } ^\text{n}\text{C}_\text{r}=\frac{\text{n}!}{\text{r}!(\text{n}-\text{t})!}\Big]$
$=\frac{\frac{5.4}{2}\times11+5}{\frac{16\times15\times14\times13}{4\times3\times2}}$
$=\frac{(110+5)}{1820}$
$=\frac{115}{1820}$
$=\frac{23}{364}$
Required probability $=\frac{23}{364}$
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Question 425 Marks
An item is manufactured by three machines $\mathrm{A}, \mathrm{B}$ and $C$ . Out of the total number of items manufactured during a specified period, $50 \%$ are manufactured on machine A, $30 \%$ on B and $20 \%$ on $C$, $2 \%$ of the items produced on A and $2 \%$ of items produced on $B$ are defective and $3 \%$ of these produced on $C$ are defective. All the items stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine $A$ ?
Answer
Consider the following events:
$E_1$ = Item is produced by machine A,
$E_2$ = Item is produced by machine B,
$E_3$ = Item is produced by machine C,
A = Item is defective
Clearly,
$\text{P}(\text{E}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{E}_2)=\frac{30}{100}=\frac{3}{10},\text{P}(\text{E}_3)=\frac{20}{100}=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{3}{100}$
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
$=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{100}}{\frac{1}{2}\times\frac{2}{100}+\frac{3}{10}\times\frac{2}{100}+\frac{1}{5}\times\frac{3}{100}}$
$=\frac{5}{11}$
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Question 435 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = the first throw results in head,
B = the last throw results in tail.
Answer
A coin is tossed thrice
Samplw space = {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = The first throw results in head
A = {HHT, HTH, HHH, HTT}
B = The last throw in tail
B = {HHT, HTT, THT, TTT}
$\text{A}\cap\text{B}=\{\text{HHT, HTT}\}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}$
$\text{P(A) }\text{P(B)}=\frac{1}{2},\frac{1}{2}$
$\text{P(A) }\text{P(B)}=\frac{1}{4}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are independent events.
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Question 445 Marks
For $\mathrm{A}, \mathrm{B}$ and $C$ the chances of being selected as the manager of a firm are in the ratio $4: 1: 2$ respectively. The respective probabilities for them to introduce a radical change in marketing strategy are $0.3,0.8$ and $0.5$. If the change does take place, find the probability that it is due to the appointment of $B$ or $C$ .
Answer
Let $E_1, E_2, E_3$ and A be event as:
$E_1$ = A is appointed
$E_2$ = B is appointed
$E_3$ = C is appointed
A = A change does take place
$\text{P}(\text{E}_1)=\frac{4}{7}$
$\text{P}(\text{E}_2)=\frac{1}{7}$
$\text{P}(\text{E}_3)=\frac{2}{7}$
$P(A|E_1)$ = P(Changes tale place by A)
= 0.3
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Changes take place by B)
= 0.8
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Changes take place by C)
= 0.5
To find, P(Changes were taken place by B or C) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)+\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)$
By baye's theorem,
$=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)+\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{7}\times\frac{8}{10}+\frac{2}{7}\times\frac{5}{10}}{\frac{4}{7}\times\frac{3}{10}+\frac{1}{7}\times\frac{8}{10}\times\frac{2}{7}\times\frac{5}{10}}$
$=\frac{\frac{18}{70}}{\frac{30}{70}}$
$=\frac{18}{30}$
$=\frac{3}{5}$
Required probability $=\frac{3}{5}$
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Question 455 Marks
An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X, 9 out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y. Calculate the probability that the assembled product will not be defective.
Answer
Given,
Prat X has 9 out of 100 defective
⇒ Part X has 91 out of 100 non defective
Part Y has out of 100 defective
⇒ Part Y has 95 out of 100 non defective
Consider,
X = A non defective part X
Y = A non defective Part Y
$\Rightarrow\ \text{P(x)}=\frac{91}{100}$ and $\text{P(Y)}=\frac{95}{100}$
= P(Assembled product will bot be defective)
= P(Niether X defective nor Y defective)
$=\text{P}(\text{X}\cap\text{Y})$
$=\text{P(X) }\text{P(Y)}$
$=\frac{91}{100}\times\frac{95}{100}$
$=0.8645$
Required probability = 0.8645
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Question 465 Marks
A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.
Answer
Given bag contains 4 white, 7 black and 5 red balls.
Total number of balls = 16
Three balls are drawn without replacement
A = First ball is white
B = Second ball is black
C = Third balls is red
P (Three balls drawn are white, black, red respectively)
$=\text{P}(\text{A}) \text{ P}\Big(\frac{\text{B}}{\text{A}}\Big) \text{ P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)$
$=\frac{4}{16}\times\frac{7}{15}\times\frac{5}{14}$
$=\frac{1}{24}$
Required probability $=\frac{1}{24}$
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Question 475 Marks
In a certain college, $4 \%$ of boys and $1 \%$ of girls are taller than $1.75$ metres. Further more, $60 \%$ of the students in the colleges are girls. A student selected at random from the college is found to be taller than $1.75$ metres. Find the probability that the selected students is girl.
Answer
Consider the following events
$E_1$ = The selected student is a girl
$E_2$ = The selected student is not a girl
A = The student is taller than 1.75 meters
We have,
$\text{P}(\text{E}_1)=60\%=\frac{60}{100}=0.6$
$\text{P}(\text{E}_2)=1-\text{P}(\text{E}_1)=1-0.6=0.4$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability that the student is taller than 1.75 meters given that the student is a girl
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{1}{100}=0.01$
And
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability that the student is taller than 1.75 meters given that the student is not a girl
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{4}{100}=0.04$
Now,
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.6\times0.01}{0.6\times0.01+0.4\times0.04}$
$=\frac{\frac{6}{1000}}{\frac{22}{1000}}$
$=\frac{3}{11}$
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Question 485 Marks
An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3 or 4 on the second toss.
Answer
Given an unbiased die is tossed twise
A = Getting 4, 5 or 6 on the first toss
B = 1, 2, 3 or 4 on second toss
$\Rightarrow\ \text{P(B)}=\frac{3}{6}$
$\text{P(A)}=\frac{1}{2}$
and, $\text{P(B)}=\frac{4}{6}$
$\text{P(B)}=\frac{2}{3}$
P (Getting 4, 5 or 6 on the first toss and 1, 2, 3 or 4 on second toss)
$=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A) }\text{P(B)}$
$=\frac{1}{2}\times\frac{2}{3}$
$=\frac{1}{3}$
Required probability $=\frac{1}{3}$
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Question 495 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = the number of heads is two,
B = the last throw results in head.
Answer
Sample space for throwing a coin thrice
= {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = The number of heads is two
A = {HHT, THH, HTH}
B = The Last throw results in head
B = {HHH, HTH, THH, TTH}
$\text{A}\cap\text{B}=\{\text{THH, HTH}\}$
$\text{P(A)}=\frac{3}{8}$
$\text{P(B)}=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}$
$\text{P(A)}.\text{P}(\text{B})=\frac{3}{8}\times\frac{1}{2}$
$=\frac{3}{16}$
$\text{P(A)}.\text{P(B)}\neq\text{P}(\text{A}\cap\text{B})$
So, A and B are not independent events.
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Question 505 Marks
A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.
Answer
A = 4 appears on third toss, if a die is thrown three times
= {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4)
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4)
(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4)
(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)
(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4)
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = 6 and 5 appears respectively on first two tosses, if die is tosses three times
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
$\text{A}\cap\text{B}=\{(6,5,4)\}$
Required probability $=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required Probability $=\frac{1}{6}$
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Question 515 Marks
Bag A contains $3$ red and $5$ black balls, while bag $B$ contains $4$ red and $4$ black balls. Two balls are transferred at random from bag $A$ to bag $B$ and then a ball is drawn from bag $B$ at random. If the ball drawn from bag $B$ is found to be red find the probability that two red balls were transferred from $A$ to $B$.
Answer
It is given that bag A contains 3 red and 5 black balls and bag B contains 4 red and 4 black balls.
Let $E_1, E_2, E_3$ and A be the events as defined below:
$E_1$ : Two red balls are transferred from bag A to Bag B.
$E_2$ : One red ball and one black ball is transferred from bag A to bag B.
$E_3$ : Two black balls are transferred from bag A to bag B.
A = Ball drawn from bag B is red.
So,
$\text{P}(\text{E}_1)=\frac{^{3}\text{C}_2}{^{8}\text{C}_2}=\frac{3}{28}$
$\text{P}(\text{E}_2)=\frac{^{3}\text{C}_1\times ^{5}\text{C}_1}{^{8}\text{C}_2}=\frac{15}{28}$
$\text{P}(\text{E}_3)=\frac{^{5}\text{C}_2}{^{8}\text{C}_2}=\frac{10}{28}$
Also,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{6}{10}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{5}{10}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{4}{10}$
$\therefore$ Required probability = Probability that two red balls were transferred from A to B given that the ball drawn from bag B is red
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
[Using Baye's Theorem]
$=\frac{\frac{3}{28}\times\frac{6}{10}}{\frac{3}{28}\times\frac{6}{10}+\frac{15}{28}\times\frac{5}{10}+\frac{10}{28}\times\frac{4}{10}}$
$=\frac{18}{18+75+40}$
$=\frac{18}{133}$
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Question 525 Marks
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big).$
Answer
Given,
$\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{3}+\frac{1}{5}-\frac{11}{30}$
$=\frac{10+6-11}{30}$
$=\frac{5}{30}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{6}}{\frac{1}{5}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{6}}{\frac{1}{3}}$
$=\frac{1}{6}\times\frac{3}{1}$
$=\frac{1}{2}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6},\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{1}{2}$
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Question 535 Marks
A couple has two children. Find the probability that both the children are,
  1. Males, if it is known that at least one of the children is male.
  2. Females, if it is known that the elder child is a female.
Answer
Consider the given events.
A = Both the children are female.
B = The elder child is a female.
C = At least one child is a male.
D = Both children are male.
Clearly,
S $= {M_1M_2, M_1F_2, F_1M_2, F_1F_2}$
A $= {F_1F_2}$
B $= {F_1M_2, F_1F_2​​​​​​​}$
C $= {M_1F_2, F_1M_2, M_1M_2}$
D $= {M_1M_2}$
[Here, first child is elder and second is younger]
$\text{D}\cap\text{C}=\big\{\text{M}_1\text{M}_2\big\}$ and $\text{A}\cap\text{B}=\big\{\text{F}_1\text{F}_2\big\}$
  1. Required probability $=\text{P}\Big(\frac{\text{D}}{\text{C}}\Big)=\frac{\text{n}(\text{D}\cap\text{C})}{\text{n}(\text{C})}=\frac{1}{3}$
  2. Required Probability $= \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$
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Question 545 Marks
By examining the chest X-ray, probability that T.B. is detected when a person is actually suffering is $0.99$ . The probability that the doctor diagnoses incorrectly that a person has T.B. on the basic of X-ray is $0.001$. In a certain city $1$ in $1000$ persons suffers from T.B. A person is selected at random is diagnosed to have T.B. what is the chance that he actually has T.B.?
Answer
Consider events $E_1, E_2$ and A as
$E_1$ = The person selected is actually having T.B.
$E_2$ = The person selected not having T.B.
$E_3$ = The person diagnosed to have T.B.
Given,
$\text{P}(\text{E}_1)=\frac{1}{1000}$
$\text{P}(\text{E}_2)=\frac{999}{1000}$
$P(A|E_1)$ = P(Person diagnosed to have T.B. and he is actually having T.B.)
$=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ P(Person diagnossed to have a T.B. and he is not a actually having T.B.)
$0.001$
To find, P(Person diagnosed to have T.B. is actually having T.B.) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{1000}\times0.99}{\frac{1}{1000}\times0.99+\frac{999}{1000}\times0.001}$
$=\frac{990}{990+999}$
$=\frac{990}{1989}$
$=\frac{110}{221}$
Required probability $=\frac{110}{221}$
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Question 555 Marks
A bag contains 20 tickets, numbered from 1 to 20. Two tickets are drawn without replacement. What is the probability that the first ticket has an even number and the second an odd number.
Answer
Total number of tickets are 20 numbered from 1, 2, 3, ..... 20.
Number of tickets with even numbers,
= 10 [Since, even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
Number of tickets with odd numbers,
= 10 [Since, odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17,19]
Two cards are drawn without replacement.
A = tickets with even numbers
B = tickets with odd numbers
P (first ticket has even number and second has odd number)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{10}{20}\times\frac{10}{19}$
$=\frac{5}{19}$
Required probability $=\frac{5}{19}$
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Question 565 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$ then find $\text{P}(\text{A}|\text{B}), \text{ P}(\text{B}|\text{A}), \text{ P}(\overline{\text{A}}|\text{B})$ and $\text{P}(\overline{\text{A}}|\overline{\text{B}}).$
Answer
We have,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$
Also, $\text{P}(\overline{\text{B}})=1-\text{P(B)}=1-\frac{1}{3}=\frac{2}{3}$
As, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$=\frac{6+4-3}{12}$
$\Rightarrow\ \text{P}(\text{A}\cup\text{B})=\frac{7}{12}$
Also, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{3}-\frac{1}{4}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{4-3}{12}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{12}$
And, $\Rightarrow\ \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$=1-\text{P}({\text{A}\cup\text{B}})$
$=1-\frac{7}{12}$
$=\frac{5}{12}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{4},$
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{2}\Big)}=\frac{2}{4}=\frac{1}{2},$
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{1}{12}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{12}=\frac{1}{4}$ and
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}=\frac{\Big(\frac{5}{12}\Big)}{\Big(\frac{2}{3}\Big)}=\frac{15}{24}=\frac{5}{8}$
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Question 575 Marks
Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.
Answer
A pair of dice are thrown. It has 36 elem ents in its samplw space.
A = Occurence of number 4 on firs die
A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
B = Occurence of 5 on second die
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
$\text{A}\cap\text{B}=\Big\{(4,5)\Big\}$
$\text{P(A)}=\frac{6}{36}=\frac{1}{6}$
$\text{P(B)}=\frac{6}{36}=\frac{1}{6}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}\frac{1}{6}\times\frac{1}{6}$
$=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are indepepndent events.
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Question 585 Marks
Coloured balls are distributed in four boxes as shown in the following table:
Box
Colour
Black
White
Red
Blue
I
II
III
IV
3
2
1
4
4
2
2
3
5
2
3
1
6
2
1
5
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.
Answer
Let $A, E_1, E_2, E_3$ and $E_4$​​​​​​​ denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{18}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_4}\Big)=\frac{4}{13}$
Using Bayes' theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{4}\times\frac{1}{7}}{\frac{1}{4}\times\frac{3}{18}+\frac{1}{4}\times\frac{2}{8}+\frac{1}{4}\times\frac{1}{7}+\frac{1}{4}\times\frac{4}{13}}$
$=\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{1}{13}}=\frac{156}{947}$
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Question 595 Marks
One bag contains $4$ yellow and $5$ red balls. Another bag contains $6$ yellow and $3$ red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.
Answer
Bag I contains 4 yellow and 5 red balls
Bag II contains 6 yellow and 3 red balls
Transfer can be done in two ways:
I - A red ball is transferred from bag I to bag II and then one yellow ball is drawn from bag II.
II - A red ball is transferred from bag I to bag II and then one yelllow ball is drawn from bag II.
Let $E_1, E_2$ and A be events as:
$E_1$ = One yellow ball drawn from bag I
$E_2$ = One red ball drawn from bag I
A = one yellow ball draw from bag II.
$\text{P}(\text{E}_1)=\frac{4}{9}$
$\text{P}(\text{E}_2)=\frac{4}{9}$
$\text{P}(\text{A}|\text{E}_1)=\frac{7}{10}$
[Since $E_1$ has increased one yellow ball in bag II]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{6}{10}$
[Since $E_2$ has increased one red ball in bag II]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{4}{9}\times\frac{7}{10}\times\frac{5}{9}\times\frac{6}{10}$
$=\frac{28+30}{90}$
$=\frac{58}{90}$
$=\frac{29}{45}$
Required probability $=\frac{29}{45}$
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Question 605 Marks
Two numbers are selected at random from integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.
Answer
Two numbers are selected at random from integers 1 through 9.

A = Both numbers are odd

A = {(3, 1), (5, 1), (7, 1), (9, 1), (3, 5), (3, 7), (9, 3), (5, 3), (5, 7), (5, 9), (7, 3), (7, 5), (7, 9), (9, 3), (9, 5), (9, 7)}

B = Sum of both numbers is even

A = Sum of both numbers is 2, 4, 6, 8, 10, 12, 14, 16 or 18 = {(1, 3), (1, 5), (2, 4), (1, 7), (2, 6), (3, 5), (1, 9), (2, 8), (3, 7), (4, 6), (7, 5), (8, 4), (9, 3), (8, 6), (9, 5), (9, 7)}

$(\text{A}\cap\text{B})=$ {(1, 3), (1, 5), (1, 7), (3, 5), (1, 9), (3, 7), (7, 5), (9, 3), (9, 5), (9, 7)}

$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$

$=\frac{10}{16}$

Required probability $=\frac{10}{16}$
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Question 615 Marks
The probability that A hits a target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$, What is the probability that the target will be hit, if each one of A and B shoots at the target?
Answer
Given,
Probability that A hits a target $=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}=\frac{1}{3}$
Probability that B hits the targer $=\frac{2}{5}$
$\Rightarrow\ \text{P(B)}=\frac{2}{5}$
P (Target will be hit)
= 1 - P (target will not be hit)
= 1 - P (Niether A non B hits the target)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})$
$=1-[1-\text{P(A)}][1-\text{P}(\overline{\text{B}})]$
$=1-\Big[1-\frac{1}{3}\Big]\Big[1-\frac{2}{5}\Big]$
$=1-\frac{2}{3},\frac{3}{5}$
$=1-\frac{2}{5}$
$=\frac{2}{5}$
Required probability $=\frac{2}{5}$
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Question 625 Marks
A bag contains $3$ white and $2$ black balls and another bag contains $2$ white and $4$ black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.
Answer
Bag I contains 3 white and 2 black balls
Bag II contains 2 white and 4 black balls
One bag is chosen at random, then one ball is drawn and its is while.
Let $E_1, E_2$ and A be events as:
$E_1$ = Selecting bag I
$E_2$ = Selecting bag II
A = Drawing one white ball
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since there are only 2 bags]
$P(A|E_1)$ = P(Drawing a white ball from bag I)
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Drawing a white ball from bag II]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{2}{6}$
$=\frac{3}{10}+\frac{2}{12}$
$=\frac{18+10}{60}$
$=\frac{28}{60}$
$=\frac{7}{15}$
Required probability $=\frac{7}{15}$
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Question 635 Marks
A and B are two independent events. The probability that A and B occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Find the probability of occurrence of two events.
Answer
Given
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{3}$
We know that,
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$\frac{1}{3}=(1-\text{P(A)})(1-\text{P(A)})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P(A) }\text{P(B)}$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P}(\text{A}\cap\text{B})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(B)}+\frac{1}{6}$
$\text{P(A)}+\text{P(A)}=\frac{1}{1}+\frac{1}{6}-\frac{1}{3}$
$=\frac{6+1-2}{6}$
$\text{P(A)}+\text{P(B)}=\frac{5}{6}$
$\text{P(A)}=\frac{5}{6}-\text{P(B)}\ .....\text{(i)}$
Given, $\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P(A) } \text{P(B)}=\frac{1}{6}$
$\Big[\frac{5}{6}-\text{P(B)}\Big]\text{P(B)}=\frac{1}{6}$
[Using equation (i)]
$\Rightarrow\ \frac{5}{6}\text{P(B)}-\big\{\text{P(B)}\big\}^2=\frac{1}{6}$
$\Rightarrow\ \{\text{P(B)}\}^2-\frac{5}{6}\text{P(B)}+\frac{1}{6}=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-5\text{P(B)}+1=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-3\text{P(B)}-2\text{P(B)}+1=0$
$\Rightarrow\ 3\text{P(B)}[2\text{P(B)}-1]-1[2\text{P(B)}-1]=0$
$\Rightarrow\ [2\text{P(B)}-1][3\text{P(B)-1}]=0$
$\Rightarrow\ 2\text{P(B)}-1 = 0\text{ or }3\text{P(B)}-1=0$
$\Rightarrow\ \text{P(B)}=\frac{1}{2} \text{ or P(B)}=\frac{1}{3}$
⇒ Using equation (i),
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}$
Hence, $\text{P(B)}=\frac{1}{2},\text{P(A)}=\frac{1}{3} \text{ or }\text{P(B)}=\frac{1}{3},\text{P(A)}=\frac{1}{2}$
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Question 645 Marks
A bag contains 3 red and 2 black balls. One ball is drawn from it at random. Its colour is noted and then it is put back in the bag. A second draw is made and the same procedure is repeated. Find the probability of drawing,
  1. Two red balls,
  2. Two black balls,
  3. First red and second black ball.
Answer
Given bag contains 3 red and 2 black balls.
A = Getting one red ball
$\Rightarrow\ \text{P(A)}=\frac{3}{5}$
B = Getting one black ball
$\Rightarrow\ \text{P(B)}=\frac{2}{5}$
  1. P(Getting two red balls)
= P(A) P(A)

$=\frac{3}{5}\times\frac{3}{5}$

$=\frac{9}{25}$

P(Getting two red balls) $=\frac{9}{25}$
  1. P(Getting two black balls)
= P(B) P(B)

$=\frac{2}{5}\times\frac{2}{5}$

$=\frac{4}{25}$

P(Getting two black balls) $=\frac{4}{25}$
  1. P(Getting first red and second black ball)
= P(A) P(B)

$=\frac{3}{5}\times\frac{2}{5}$

$=\frac{6}{25}$

P(Getting first red and second black ball) $=\frac{6}{25}$
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Question 655 Marks
A die is thrown thrice. Find the probability of getting an odd number at least once.
Answer
P(getting an odd number in one throw) $=\frac{1}{2}$
Here, getting an odd number in three throws refers to 3 independent events.
$\text{P(A)}=\text{P{B}}=\text{P(C)}=\frac{1}{2}$
$\text{P}(\text{A}\cup\text{B}\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)} \\ -[\text{P}(\text{A}\cap\text{B})+(\text{B}\cap\text{C})+(\text{C}\cap\text{A})]+\text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\Big[\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\Big] \\ +\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{2}-\frac{3}{4}+\frac{1}{8}$
$=\frac{12-6+1}{8}$
$=\frac{7}{8}$
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Question 665 Marks
A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.
Answer
A pair of dice is thrownA = getting sum 8 or more
= Getting sum 8, 9, 10, 11 or 12 on the pair of dice
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6)
(4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4)
(5, 6), (6, 5), (6, 6)
B = 4 on first die
B = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
$(\text{A}\cap\text{B})=\{(4, 4), (4, 5), (4, 6)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{6}$
$=\frac{1}{2}$
Required probability $=\frac{1}{2}$
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Question 675 Marks
Suppose a girl throws a die. If she gets $1$ or $2$, she tosses a coin three times and notes the number of tails. If she gets $3, 4, 5$ or $6$, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', then what is the probability that she threw $3, 4, 5$ or $6$ with the die?
Answer
Let $E_1$ be the event that the outcome on the die is $1$ or $2$ and $E_2$​​​​​​​ be the event that the outcome on the die is $3, 4, 5$ or $6$. then,
$\text{P}(\text{E}_1)=\frac{2}{6}=\frac{1}{3}$ and $\text{P}(\text{E}_2)=\frac{4}{6}=\frac{2}{3}$
Let A bethe event of getting exaxtly one 'tail'.
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)$ = Probability of getting exactly one tail by tossing the coin three times if she tesd $1$ or $2$ = $P(A|E_2)$ = Probability of getting exactly one tail in a single throw of a coin ifshe gets $3, 4, 5$ or $5$ = As, the probability that the girlthrew $3, 4, 5$ or $6$ with the die, if she obtained exaxtly one tail, is given by $P(E_2|A)$.
​​​​​​​So, by using Baye's therorem, we get
$\text{P}(\text{E}_2|\text{A})=\frac{\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)}$
$=\frac{\Big(\frac{2}{3}\times\frac{1}{2}\Big)}{\Big(\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}\Big)}$
$=\frac{\Big(\frac{2}{6}\Big)}{\Big(\frac{1}{8}+\frac{1}{3}\Big)}$
$=\frac{\Big(\frac{2}{6}\Big)}{\Big(\frac{11}{24}\Big)}$
$=\frac{24\times2}{11\times6}$
$=\frac{8}{11}$
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Question 685 Marks
Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.
Answer
Let $E_1$ and $E_2$ denote the events that the first group and the second group win the competition, respectively. Let A be the event of introducing a new product.
$P(E_1)$ = Probability that the first group wins the competition = 0.6
$P(E_2)$ = Probability that the second group wins the competition = 0.4
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability of introducing a new product if the first group wins = 0.7
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability of introducing a new product if the second group wins = 0.3
The probability that the new product is introduced by the second group is given by $\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big).$
Using Bayes' theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}$
$=\frac{0.12}{0.54}=\frac{2}{9}$
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Question 695 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ then find $\text{P}(\overline{\text{A}}|\text{B}).$
Answer
We have,
$\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$
As, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{9}{13}-\frac{4}{13}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{9-4}{13}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{5}{13}$
Now,
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{5}{13}\Big)}{\Big(\frac{9}{13}\Big)}=\frac{5}{9}$
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Question 705 Marks
There are three urns $A, B,$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$ contains $4$ red and $4$ black balls. One ball is drawn from each of these urns. What is the probability that $3$ balls drawn consists of $2$ red balls and a black ball?
Answer
Urn $A$ contains $4$ red $(R_1)$ and $3$ black $(B_1)$ balls
Urn $B$ contains $5$ red $(R_2)$ and $4$ black $(B_2)$ balls
Urn $C$ contains $4$ red $(R_3)$ and $4$ balck $(B_3)$ balls.
$P$ ($3$ balls drawn consists or $2$ red and a black ball)
$=\text{P}\big[(\text{R}_1\cap\text{R}_2\cap\text{R}_3)\cup(\text{R}_1\cap\text{B}_2\cap\text{R}_3)\cap(\text{B}_1\cap\text{R}_2\cap\text{R}_3)\big]$
$=\text{P}(\text{R}_1\cap\text{R}_2\cap\text{R}_3)+\text{P}(\text{R}_1\cap\text{B}_2\cap\text{R}_3)+\text{P}(\text{B}_1\cap\text{R}_2\cap\text{R}_3)$
$=\text{P}(\text{R}_1)\text{P}(\text{R}_2)\text{P}(\text{R}_3)+\text{P}(\text{R}_1)\text{P}(\text{B}_2)\text{P}(\text{R}_3)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)\text{P}(\text{R}_3)$
$=\frac{4}{7}\times\frac{5}{9}\times\frac{4}{8}+\frac{4}{7}\times\frac{4}{9}\times\frac{4}{8}+\frac{3}{7}\times\frac{5}{9}\times\frac{4}{8}$
$=\frac{80+64+60}{504}$
$=\frac{204}{504}$
$=\frac{17}{42}$
Required probability $=\frac{17}{42}$
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Question 715 Marks
In a group of $400$ people, $160$ are smokers and non-vegetarian, $100$ are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are $35\%, 20\%$ and $10\%$ respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?
Answer
Let $E_1, E_2, E_3$​​​​​​​ be the events that the people are smokers and non-vegetarian, skokers and vegetarian, and non-smokers and vegetarian respectively.
$\text{P}(\text{E}_1)=\frac{2}{5},\text{P}(\text{E}_2)=\frac{1}{4},\text{P}(\text{E}_1)=\frac{7}{20}$
Let A denote the event that the person has the special chest disease. It is given that
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=0.35,\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=020,\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=0.10$
We have to find $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{2}{5}(0.35)}{\frac{2}{5}(0.35)+\frac{1}{4}(0.20)+\frac{7}{20}(0.10)}=\frac{\frac{7}{50}}{\Big(\frac{7}{50}\Big)+\Big(\frac{1}{20}\Big)+\Big(\frac{7}{200}\Big)}$
$=\frac{\frac{7}{50}}{\frac{9}{40}}=\frac{28}{45}$
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Question 725 Marks
Suppose $5$ men out of $100$ and $25$ women out of $1000$ are good orators. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.
Answer
Given,
5 man out of 100 and 25 women out of 1000 are good orators.
Consider $E_1, E_2$ and A events as:
$E_1$ = Selected persom is male
$E_2$ = Selected person is famale
$E_3$ = Selected person is an orator
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since number of ,ales and females are equal]
$P(A|E_1)$ = P(Selecting a male orator)
$=\frac{5}{100}$
$=\frac{1}{20}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting a female orator)
$=\frac{25}{1000}$
$=\frac{1}{40}$
To find, P(Prator selected is a male) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{1}{20}}{\frac{1}{2}\times\frac{1}{20}+\frac{1}{2}\times\frac{1}{40}}$
$=\frac{\frac{1}{40}}{\frac{1}{40}+\frac{1}{80}}$
$=\frac{1}{40}\times\frac{80}{3}$
$=\frac{2}{3}$
Required probability $=\frac{2}{3}.$
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Question 735 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that,
  1. Both balls are red,
  2. First ball is black and second is red,
  3. One of them is black and other is red.
Answer
The Box contains 10 black balls and 8 red balls.
Then $\text{P(Black ball)}=\frac{10}{18}$
$\text{P(red ball)}=\frac{8}{18}$
  1. P(Both ballls are red) $=\frac{8}{18}\times\frac{8}{18}=\frac{16}{81}$
  2. P (First ball is black and second is red) $=\frac{10}{18}\times\frac{8}{18}=\frac{20}{81}$
  3. P (One of them is black and other is red)
$=\frac{10}{18}\times\frac{8}{18}+\frac{8}{18}\times\frac{10}{18}$

$=2\Big(\frac{20}{81}\Big)$

$=\frac{40}{81}$
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Question 745 Marks
Three persons $A, B$ and $C$ apply for a job of Manager in a Private Company. Chances of their selection ($A, B$ and $C$) are in the ratio $1 : 2 : 4$. The probabilities that $A, B$ and $C$ can introduce changes to improve profits of the company are $0.8, 0.5$ and $0.3$, respectively. If the change does not take place, find the probability that it is due to the appointment of $C$.
Answer
Let $E_1, E_2$ and $E_3$ be the events denoting the selecting of A, B and C as managers, respectively.
$P(E_1)$ = Ptobability of selection of A $=\frac{1}{7}$
$P(E_2)$ = Probability of selection of B $=\frac{2}{7}$
$P(E_3)$ = Probability of selection of C $=\frac{4}{7}$
Let be the event denoting the change not taking place.
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability that A does not introduce change = 0.2
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability that B does not introduce change = 0.5
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=$ Probability that C does not introduce change = 0.7
$\therefore$ Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By Baye's theorem, we have
$\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{4}{7}\times0.7}{\frac{1}{7}\times0.2+\frac{2}{7}\times0.5+\frac{4}{7}\times0.7}$
$=\frac{2.8}{0.2+1+2.8}$
$=\frac{2.8}{4}=0.7$
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Question 755 Marks
In a factory, machine A produces $30 \%$ of the total output, machine B produces $25 \%$ and the machine C produces the remaining output. If defective items produced by machines $A , B$ and C are $1 \%, 1.2 \%, 2 \%$ respectively. Three machines working together produce $10000$ items in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine B?
Answer
Consider events $E _1, E _2, E _3$ and Aas:
$E _1=$ Selecting product from machine A
$E_2=$ Selecting product from machine $B$
$E_3=$ Selecting product from machine $C$
$A=$ Selecting a standard quality product
$\text{P}(\text{E}_1)=\frac{30}{100}$
$\text{P}(\text{E}_2)=\frac{25}{100}$
$\text{P}(\text{E}_3)=\frac{45}{100}$
P($A|E_1$​​​​​​​) = P(Selecting defective product from machine A)
$=\frac{1}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting defective prodcut from machine B)
$=\frac{1.2}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Selecting defective product from machine C)
$=\frac{2}{100}$
To find, P(Selecting defective product is produced by machine B)
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{25}{100}\times\frac{12}{1000}}{\frac{30}{100}\times\frac{1}{100}+\frac{25}{100}\times\frac{12}{1000}+\frac{45}{100}\times\frac{2}{100}}$
$=\frac{300}{300+300+900}$
$=\frac{300}{1500}$
$=\frac{1}{5}$
Required probability $=\frac{1}{5}$
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Question 765 Marks
A bag contains $4$ white and $5$ black balls and another bag contains $3$ white and $4$ black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.
Answer
There are two bags.
Bag (1) contain 4 white and 5 black balls.
Bag (2) contain 4 white and 4 black balls.
A ball is taken from bag (i) and without seeing its colout is put in second bag. Then a ball is drawn from bag 2 and is white in colour.
A (White ball from bag 1) $=\frac{4}{9}$
$\text{P}(\text{W}_1)=\frac{4}{9}$
P(Black ball from bag 1) $=\frac{5}{9}$
$\text{P}(\text{B}_1)=\frac{5}{9}$
P(white ball from bag 2 given $B_1$​​​​​​​ transfer)
$\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)=\frac{3}{8}$
P(White from bag 2 given $W_1$ transfer)
$\text{P}\Big(\frac{\text{W}_2}{\text{W}_2}\Big)=\frac{4}{8}$
$=\frac{1}{2}$
P(white from bag 2)
$=\text{P}(\text{B}_1)\text{P}\Big(\frac{\text{W}_2}{\text{B}_1}\Big)+\text{P}(\text{W}_1)\text{P}\Big(\frac{\text{W}_2}{\text{W}_1}\Big)$
$=\frac{5}{8}\times\frac{3}{8}+\frac{4}{9}\times\frac{1}{2}$
$=\frac{15}{72}+\frac{4}{18}$
$=\frac{31}{72}$
Required probability $=\frac{31}{72}$
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Question 775 Marks
The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a trouser is 0.3, and the probability that he will buy a shirt given that he buys a trouser is 0.4. Find the probability that he will buy both a shirt and a trouser. Find also the probability that he will buy a trouser given that he buys a shirt.
Answer
Given,
Probability that a person buys a shirt (S) = P(S) = 0.2
Probability that he buys a trouser (T) = P(T) = 0.3
$\text{P}\Big(\frac{\text{S}}{\text{T}}\Big)=0.4$
We know that,
$\text{P}\Big(\frac{\text{S}}{\text{T}}\Big)=\frac{\text{P}(\text{S}\cap\text{T})}{\text{P}(\text{T})}$
$0.4=\frac{\text{P}(\text{S}\cap\text{T})}{0.3}$
$\text{P}(\text{S}\cap\text{T})=0.4\times0.3$
$\text{P}(\text{S}\cap\text{T})=0.12$
Probability that he buys a shirt and a trouser both = 0.12
$\text{P}\Big(\frac{\text{T}}{\text{S}}\Big)=\frac{\text{P}(\text{S}\cap\text{T})}{\text{P}(\text{S})}$
$=\frac{0.12}{0.2}$
$\text{P}\Big(\frac{\text{T}}{\text{S}}\Big)=\frac{12}{20}$
$=\frac{3}{5}$
$=0.6$
Probability that he buys a trouser given that he buys a shirt = 0.6
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Question 785 Marks
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from,
  1. Box A?
  2. Box B?
  3. Box C?
Answer
$\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big),\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big),\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)=\frac{\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)\text{P(A)}}{\text{P}\Big(\frac{\text{A}}{\text{Red}}\Big)\text{P(A)}+\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big)\text{P(B)}+\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)\text{P(C)}+\text{P}\Big(\frac{\text{D}}{\text{Red}}\Big)\text{P(D)}}$
$=\frac{\frac{1}{10}\times\frac{1}{4}}{\frac{1}{10}\times\frac{1}{4}+\frac{6}{10}\times\frac{1}{4}+\frac{8}{10}\times\frac{1}{4}+0}$
$=\frac{1}{1+6+8}=\frac{1}{15}$
Similarly,
$\text{P}\Big(\frac{\text{B}}{\text{Red}}\Big)=\frac{6}{15}$
$\text{P}\Big(\frac{\text{C}}{\text{Red}}\Big)=\frac{8}{15}$
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Question 795 Marks
The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that,
  1. At least one of the events will occur,
  2. None of the events will occur.
Answer
Given,
The adds against a certain event (say, A) are 5 to 2
$\Rightarrow\ \text{P}(\overline{\text{A}})=\frac{5}{5+2}$
$\text{P}(\overline{\text{A}})=\frac{5}{7}$
The odds in favour of another event (say, B) are 6 to 5
$\Rightarrow\ \text{P(B)}=\frac{6}{5+6}$
$\text{P(B)}=\frac{6}{11}$
$\text{P}(\overline{\text{B}})=1-\frac{6}{11}$
$\text{P}(\overline{\text{B}})=\frac{5}{11}$
  1. P (At least one of the events will occur)
= 1 - P (None of events occur)

$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

$=1-\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$

[Since events are independent]

$=1-\frac{5}{7}\times\frac{5}{11}$

$=1-\frac{25}{77}$

$=\frac{52}{77}$

Required probability $=\frac{52}{77}$
  1. P (None of the events will occur)
$=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

$=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$

$=\frac{5}{7}\times\frac{5}{11}$

$=\frac{25}{77}$

Required probability $=\frac{25}{77}$
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Question 805 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{1}{3},\text{P(B)}=\frac{1}{4}$ and $\text{P}(\text{A}\cap\text{B})=\frac{5}{12},$ then find P(A|B) and P(B|A).
Answer
We have,
$\text{P(A)}=\frac{1}{3},\text{P(B)}=\frac{1}{4}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{12}$
As, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{3}+\frac{1}{4}-\frac{5}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{2}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{1}{6}\Big)}{\Big(\frac{1}{4}\Big)}=\frac{4}{6}=\frac{2}{3}$ and
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}=\frac{\Big(\frac{1}{6}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{6}=\frac{1}{2}$
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Question 815 Marks
A purse contains $2$ silver and $4$ copper coins. A second purse contains $4$ silver and $3$ copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin?
Answer
Purse (I) contains (2) silver and 4 copper coins
Purse (II) contains 4 silver and 3 copper coins
Let $E_1, E_2 $and A are defined as
$E_1 =$ Selecting purse I
$E_2 =$ Selectin purse II
A = Drawinf a silver coin
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since, there are only 2 purses]
P($A|E_1$​​​​​​​) = P(A|silver coin from purse I)
$=\frac{2}{6}$
$​​=\frac{1}{3}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (A|silver coin from purse II)
By the law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{1}{3}+\frac{1}{2}\times\frac{4}{7}$
$=\frac{1}{6}+\frac{4}{14}$
$=\frac{7+12}{42}$
$=\frac{19}{42}$
Required probability $=\frac{19}{42}$
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Question 825 Marks
A and B take turns in throwing two dice, the first to throw $10$ being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio $12 : 11.$
Answer
Let E be the events of throwing 10 on a pair of dice,
E = {(4, 6), (5, 5), (6, 4)}
$\text{P(E)}=\frac{3}{37}$
$\text{P(E)}=\frac{1}{12}$
$\text{P}(\overline{\text{E}})=\frac{11}{12}$
A wins the game in first or $3^{rd}$ or $5^{th}$​​​​​​​ throw, .....
Probability that A wins in first throw $\text{P(E)}=\frac{1}{12}$
Probability that A wins in $3^{rd} $throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{11}{12}\Big)^2\Big(\frac{1}{12}\Big)$
Probability that A wins in$ 5^{th}​​​​​​​$​​​​​​​ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{11}{12}\Big)^4\Big(\frac{1}{12}\Big)$
Hence,
Probability of winning A
$=\frac{1}{12}+\Big(\frac{11}{12}\Big)^2\Big(\frac{1}{12}\Big)+\Big(\frac{11}{12}\Big)^4\Big(\frac{1}{12}\Big)$
$=\frac{1}{12}\Big[1+\Big(\frac{11}{12}\Big)^2+\Big(\frac{11}{12}\Big)^4+\ .....\Big]$
$=\frac{1}{12}\bigg[\frac{1}{1-\big(\frac{11}{12}\big)^2}\bigg]\Big[\text{Since S}_{\infty}=\frac{\text{a}}{1-\text{r}}\ \text{for G. P.}\Big]$
$=\frac{1}{12}\bigg[\frac{1}{1-\frac{121}{144}}\bigg]$
$=\frac{1}{12}\times\frac{144}{23}$
$=\frac{12}{23}$
Probability of winning B
= 1 - P(Winning A)
$=1-\frac{12}{23}$
$=\frac{11}{23}$
Chances of winning A and B are $\frac{12}{23}$ and $\frac{11}{23}$ respectively or in 12 : 11.
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Question 835 Marks
The probability that a student selected at random from a class will pass in Mathematics is $\frac{4}{5}$, and the probability that he/ she passes in Mathematics and Computer Science is $\frac{1}{2}$. What is the probability that he/ she will pass in Computer Science if it is known that he/ she has passed in Mathematics?
Answer
Given,
Probability to pass mathen atics (M)
$\text{P(M)}=\frac{4}{5}$
Probability to pass in mathematics (M) and computer Science (C)
$\text{P}(\text{M}\cap\text{C})=\frac{1}{2}$
To find, $\text{P}\Big(\frac{\text{C}}{\text{M}}\Big)$
We know tht,
$\text{P}\Big(\frac{\text{C}}{\text{M}}\Big)=\frac{\text{P}(\text{M}\cap\text{C})}{\text{P(M)}}$
$=\frac{\frac{1}{2}}{\frac{4}{5}}$
$=\frac{1}{2}\times\frac{5}{4}$
$=\frac{8}{5}$
Required probability $=\frac{8}{5}$
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Question 845 Marks
If A and B are two events such that $2\text{P(A)}=\text{P(B)}=\frac{5}{13}$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5}$ find $\text{P}(\text{A}\cap\text{B}).$
Answer
Given,
$2\text{P(A)}=\text{P(B)}=\frac{5}{13}$
$2\text{P(A)}=\frac{5}{13}$
$\Rightarrow \text{P(A)}=\frac{5}{26}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\frac{2}{5}=\frac{\text{P}(\text{A}\cap\text{B})}{\frac{5}{13}}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{5}\times\frac{5}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{13}$
We know that,
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$
$=\frac{5+10-4}{26}$
$=\frac{11}{26}$
$\text{P}(\text{A}\cap\text{B})=\frac{11}{26}$
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Question 855 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ then find $\text{P}(\text{A}\cap\text{B}),$ P(A|B) and P(B|A).
Answer
We have,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
As, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{6+5-7}{11}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{4}{11}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{b})}=\frac{\Big(\frac{4}{11}\Big)}{\Big(\frac{5}{11}\Big)}=\frac{4}{5}$ and
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\Big(\frac{4}{11}\Big)}{\Big(\frac{6}{11}\Big)}=\frac{4}{6}=\frac{2}{3}$
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Question 865 Marks
Two dice are thrown together and the total score is noted. The event E, F and G are "a total of 4", "a total of 9 or more", and "a total divisible by 5", respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.
Answer
$\text{S}=\begin{Bmatrix} (1,1),(1,2), (1, 3), (1, 4), (1, 5), (1, 6), \\ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \\ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),\$4,1), (4,2), (4,3), (4,4), (4,5), (4, 6), \$5, 1),(5,2), (5,3), (5,4), (5,5), (5,6),\$6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\end{Bmatrix}$n(S)=36
E be the event of geting a total of 4.
E = {(1, 3), (3, 1), (2, 2)}
n(E) = 3
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{36}=\frac{1}{12}$
F be event of geting a total of 9 or more.
F = {(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
n(F) = 10
$\text{P(F)}=\frac{\text{n(F)}}{\text{n(S)}}=\frac{10}{36}=\frac{5}{18}$
G be the event of getting a total divisible by 5.
G = {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)}
n(G) = 7
$\text{P(G)}=\frac{\text{n(G)}}{\text{n(S)}}=\frac{7}{36}$
No pair is independent.
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Question 875 Marks
A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. In which of the following cases are the events A and B independent?
A = the card drawn is a spade,
B = the card drawn in an ace.
Answer
A card is drawn from a pack of 52 cards
There are 13 speades and 4 Ace in which one card is ace of spade
A = The card drawn is spade
$\text{P(A)}=\frac{13}{52}$
$\text{P(A)}=\frac{1}{4}$
B = The card drawn is an ace
$\text{P(B)}=\frac{4}{52}$
$\text{P(B)}=\frac{1}{13}$
$\text{A}\cap\text{B}=$ The card drawn is an ace of spade
$\text{P}(\text{A}\cap\text{B})=\frac{1}{52}$
$\text{P(A) }\text{P(B)}=\frac{1}{4}\times\frac{1}{13}$
$=\frac{1}{52}$
$\text{P(A) }\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
Hence, A and B are independent events.
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Question 885 Marks
If P(A) = 0.4, P(B) = 0.8, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6$. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}(\text{A}\cap\text{B}).$
Answer
Given:
P(A) = 0.4, P(B) = 0.8, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6$
We know that,
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.6=\frac{\text{P}(\text{A}\cap\text{P})}{0.4}$
$\text{P}(\text{A}\cap\text{B})=0.6\times0.4$
$\text{P}(\text{A}\cap\text{B})=0.24$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.24}{0.8}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.4+0.8-0.24$
$\text{P}(\text{A}\cap\text{B})=0.96$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3, \text{P}(\text{A}\cap\text{B})=0.96$
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Question 895 Marks
An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting,
  1. 2 red balls,
  2. 2 blue balls,
  3. One red and one blue ball.
Answer
Given, Urn contains 4 red and 7 black balls.
Two balls drawn at random with replacement.
Consider,
R = Getting one red ball from urn.
$\text{P(R)}=\frac{4}{11}$
B = Getting one blue ball from urn.
$\text{P(B)}=\frac{7}{11}$
  1. P(Getting 2 red balls)
= P(R) P(R)

$=\frac{4}{11}\times\frac{4}{11}$

$=\frac{16}{121}$

Required probability $=\frac{16}{121}$
  1. P(Getting two blue balls)
= P(B) P(B)

$=\frac{7}{11}\times\frac{7}{11}$

$=\frac{49}{121}$

Required probability $=\frac{49}{121}$
  1. P(Getting one red and one blue ball)
= P(R) P(B) + P(B) P(R)

$=\frac{4}{11}\times\frac{7}{11}+\frac{7}{11}\times\frac{4}{11}$

$=\frac{28}{121}+\frac{28}{121}$

$=\frac{56}{121}$
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Question 905 Marks
Mother, father and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P (A/B) and P (B/A).
Answer
Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw
Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), ...... (6, 1, 4), (6, 2 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
Now,
$(\text{A}\cap\text{B})=\{(6, 5, 4)\}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{6}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{1}{36}$
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Question 915 Marks
A pair of dice is thrown. Let E be the event that the sum is greater than or equal to 10 and F be the event "5 appears on the first-die". Find $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)$. If F is the event "5 appears on at least one die", find $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)$.
Answer
A pair of die is thrown E = Sum is greater than or equal to 10 = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 6)}Case I:
F = 5 appears on first die = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} $\text{E}\cap\text{F}=\{(5, 5), (5, 6)\}$ $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{\text{n}(\text{E}\cap\text{F})}{\text{n}(\text{F})}$ $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{1}{3}$Case: II
F = 5 appears on at least one die = {(1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} $\text{E}\cap\text{F}=\{(5,5), (5, 6), (6, 5)\}$ $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{\text{n}(\text{E}\cap\text{F})}{\text{n}(\text{F})}$ $=\frac{3}{11}$ $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{3}{11}$
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Question 925 Marks
A test for detection of a particular disease is not fool proof. The test will correctly detect the disease $90 \%$ of the time, but will incorrectly detect the disease $1 \%$ of the time. For a large population of which an estimated $0.2 \%$ have the disease, a person is selected at random, given the test, and told that he has the disease. What are the chances that the person actually have the disease?
Answer
Consider events $E _1, E _2$ and A as:
$E _1=$ The selected person actually has disease
$E_2=$ The selected person actually has no disease
$A=$ Selected person has disease
$=\text{P}(\text{E}_1)=\frac{0.2}{100}$
$=\frac{2}{1000}$
$\text{P}(\text{E}_1)=\frac{998}{1000}$
$\text{P}(\text{A}|\text{E}_1)=\frac{90}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{100}$
To find, P(Person has disease is actually diseased) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye,s theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{2}{1000}\times\frac{90}{100}}{\frac{2}{1000}\times\frac{90}{100}+\frac{998}{1000}\times\frac{1}{100}}$
$=\frac{180}{180+998}$
$=\frac{180}{1178}$
$=\frac{90}{589}$
Required probability $=\frac{90}{589}$
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Question 935 Marks
There are three coins. One is two headed coin, another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer
Let $E_1, E_2$_ and $E_3$_ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
$\therefore\ \text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\text{P}(\text{E}_3)=\frac{1}{3}$
Let A be the event that the coin shows heads.
A two-headed coin will always show heads.
$\therefore$ P($A|E_1$​​​​​​​​​​​​​​) = P (coin showing heads, given that it is a two-headed coin) $= 1$
Probability of heads coming up, given that it is a biased coin $= 75\%$
$\therefore$ P($A|E_2​​​​​​​$​​​​​​​) = P(coin showing heads, given that it is a biased coin) $=\frac{75}{100}=\frac{3}{4}$
Since the third coin is unbiased, the probability that it shows heads is always $\frac{1}{2}$.
$\therefore$ P($A|E_3$​​​​​​​​​​​​​​) = P(coin showing heads, given that it is an unbiased coin) $=\frac{1}{2}$
The probability that the coin is two-headed, given that it shows heads, is given by $P(E_1|A).$
By using Bayes' theorem, we obtain
$=\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\times\text{P}(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}$
$=\frac{\frac{1}{3}}{\frac{1}{3}\Big(1+\frac{3}{4}+\frac{1}{2}\Big)}$
$=\frac{1}{\frac{9}{4}}$
$=\frac{4}{9}$
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Question 945 Marks
There are three categories of students in a class of 60 students:
A : Very hardworking
B : Regular but not so hardworking
C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C.
It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
Answer
Let E denote the event that the student could not get good marks in the examination.
Also, A : The event that student is very hardworking
B : The event that student is regular but not so hardworking
C : The event that student is careless and irregular
$\therefore\ \text{P(A)}=\frac{10}{60},\text{P(B)}=\frac{30}{60} \text{ and P(C)}=\frac{20}{60}$
Also,
$\text{P}\Big(\frac{\text{E}}{\text{A}}\Big)=$ Probability that the student of category A could not get good marks in the examination = 0.002
$\text{P}\Big(\frac{\text{E}}{\text{B}}\Big)=$ Probability that the student of category B could not get good marks in the examination = 0.02
$\text{P}\Big(\frac{\text{E}}{\text{C}}\Big)=$ Probability that the student of category C could not get good marks in the examination = 0.2
$\therefore$ Required probability $=\text{P}\Big(\frac{\text{C}}{\text{E}}\Big)=\frac{\text{P}(\text{C})\text{P}\Big(\frac{\text{E}}{\text{C}}\Big)}{\text{P}(\text{A})\text{P}\Big(\frac{\text{E}}{\text{A}}\Big)+\text{P}(\text{B})\text{P}\Big(\frac{\text{E}}{\text{B}}\Big)+\text{P}(\text{C})\text{P}\Big(\frac{\text{E}}{\text{C}}\Big)}$
$=\frac{\frac{20}{60}\times0.2}{\frac{10}{60}\times0.002+\frac{30}{60}\times0.02+\frac{20}{60}\times0.2}$
$=\frac{4}{4.62}=\frac{400}{462}=\frac{200}{231}$
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Question 955 Marks
Find the probability that the sum of the numbers showing on two dice is 8, given that at least one die does not show five.
Answer
Two dice are thrown
A = Sum of the numbers on dice is 8
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
B = At least one die does not show five
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
$(\text{A}\cap\text{B})=\{(2, 6), (4, 6), (6, 2)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{25}$
Require probability $=\frac{3}{25}$
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Question 965 Marks
A company has two plants to manufacture bicycles. The first plant manufactures $60 \%$ of the bicycles and the second plant $40 \%$. Out of the $80 \%$ of the bicycles are rated of standard quality at the first plant and $90 \%$ of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.
Answer
Consider events $E _1, E _2, E _3$ and Aas:
$E _1=$ Selecting bicycle from first plant
$E_2=$ Selecting bicycle from second plant
A = Selecting a standard quality bicycle
$\text{P}(\text{E}_1)=\frac{60}{100}$
$\text{P}(\text{E}_1)=\frac{40}{100}$
P($A|E_1$) = P(Selecting standard quality bicycle from firs plant)
$=\frac{80}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Selecting standard quality bicycle from second plant)
$=\frac{90}{100}$
To find, P(Selected standard quality buicycle is from second plant) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{40}{100}\times\frac{90}{100}}{\frac{60}{100}\times\frac{80}{100}+\frac{40}{100}\times\frac{90}{100}}$
$=\frac{3600}{4800+3600}$
$=\frac{3600}{8400}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
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Question 975 Marks
A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible.
What is the probability that the letter has come from,CLIFTON?
Answer
Consider events $E_1, E_2$ and A events As:
$E_1$ = Letters come from LONDON
$E_2$​​​​​​​ = Letters come from CLIFTON.
 $E_3$​​​​​​​ = Two consecutive letters visible on the envelope are on $\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$ [Since letters came either from LONDON or CLIFTON] $P(A | E_1​​​​​​​)$ = P(Two consecutive letters ON from LONDON)
$=\frac{2}{5}$ [Since LONDON has $2$ - ON and $5$ letters we consider one 'ON' as one letter]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Two consecutive letters On from CLIFTON)
$=\frac{1}{6}$ [Since CLIFTON has one 'ON' nad $6$ letters considering ON as one letter]
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{1}{6}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}$
$=\frac{\frac{1}{12}}{\frac{2}{10}+\frac{1}{12}}$
$=\frac{1}{12}\times\frac{60}{17}$
$=\frac{5}{17}$
Required probability $=\frac{5}{17}$
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Question 985 Marks
If A and B are independent events such that P(A) = p, P(B) = 2p and P(Exactly one of A and B occurs) $=\frac{5}{9},$ then find the value or p.
Answer
As, A and B are independent events.
So, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}\times\text{P(B)}\ .....(\text{i})$
Now,
P(Exactly one of A and B occurs) $=\frac{5}{9}$
$\Rightarrow\ \text{P(Only A)}+\text{P(Only B)}=\frac{5}{9}$
$\Rightarrow\ \big[\text{P(A)}-\text{P}(\text{A}\cup\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cup\text{B})\big]=\frac{5}{9}$
$\Rightarrow\ \big[\text{P(A)}-\text{P}(\text{A})+\text{P}(\text{B})\big]\times\big[\text{P(B)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]=\frac{5}{9}$
$\Rightarrow\ \text{P(A)}\times\big[1-\text{P}(\text{B})\big]+\text{P(B)}\big[1-\times\text{P}(\text{A})\big]=\frac{5}{9}$
$\Rightarrow\ \text{p}\big[1-2\text{p}\big]+2\text{p}\big[1-\text{p}\big]=\frac{5}{9}$
$=\text{p}-2\text{p}^2+2\text{p}-2\text{p}^2=\frac{5}{9}$
$\Rightarrow\ 3\text{p}-4\text{p}^2=\frac{5}{9}$
$\Rightarrow\ 27\text{p}-36\text{p}^2=5$
$\Rightarrow\ 36\text{p}^2-27\text{p}+5=0$
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