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MATHS 5 Marks Questions

Probability · MP Board (MPBSE) STD 12 Science (Madhya Pradesh - English Medium). 98 questions.

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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 $d_1, 2100$ has disease $d _2$, and others had disease $d _3 .1500$ patients with disease $d _1, 1200$ patients with disease $d _2$, 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,
⇒ 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 $1^{st}$ or $4^{th}, 7^{th} .....$ throw
A wins in first throw $\text{P(E)}=\frac{1}{6}$
A wins is $4^{th}$ throw if he fails in $1^{st}$, B fails in $2^{nd}$, C fails in $3^{rd}$ throw.
Probability of winning A in $4^{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 $2^{nd}$ or 5^{th} or $8^{th} ......$ throw.
B wins in $2^{nd}$ throw $=\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{5}{6}\Big)\Big(\frac{1}{6}\Big)$
B wins in $5^{th}$ throw if a fails in first, B fails in $2^{nd},$ C fails in $3^{rd}$, A fails in $4^{th}.$
Probabiliy of winning B in $5^{th}$ throw
$=\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P}(\overline{\text{E}})\text{P(E)}$
$=\Big(\frac{5}{4}\Big)^4\Big(\frac{1}{6}\Big)$
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 $E_1, E_2, E_3$ and A be events as:
$E_1 =$ Selecting unr I
$E_2 =$ Selecting urn II
$E_3 =$ Selecting urn III
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:
$E_1 =$ Selecting urn I
$E_2 =$ Selecting urn II
$E_3 =$ 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 X, 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 A, 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 A, 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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5 Marks Questions - MATHS STD 12 Science Questions - Vidyadip