Question 2015 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?
AnswerConsider 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 $=\frac{90}{589}$
View full question & answer→Question 2025 Marks
There are two bags, one of which contains $3$ black and $4$ white balls while the other contains $4$ black and $3$ white balls. $A$ die is thrown. If it shows up $1$ or $3,$ a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.
AnswerGiven that,
Bag $I = (3$ black $,4$ white balls$),$
Bag $II = (4$ black $,3$ white ball$),$
Let $E_1$ be the event that bag $I$ is selected and $E_2$ be the event that bag $II$ is selected
Let $E_3$ be the event that black ball is chosen.
$\therefore\text{P}(\text{E}_1)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$ and $\text{P}(\text{E}_2)=1-\frac{1}{3}=\frac{2}{3}$
And $\therefore\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)=\frac{3}{7}$ and $\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)=\frac{4}{7}$
$\therefore\text{P}(\text{E}_3)=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)$
$=\frac{1}{3}\cdot\frac{3}{7}+\frac{2}{3}\cdot\frac{4}{7}=\frac{11}{21}$
View full question & answer→Question 2035 Marks
An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.
AnswerBinomial Distribution formula is given by
$\text{P(X)}=\text{ }^{\text{n}}\text{c}_{\text{x}}\text{p}^{\text{x}}\text{q}^{\text{n}-\text{x}},$ where $\text{X}=0,1,2,\dots\text{n}$
Let X = No. of heads in a toss
We need probability of 6 or more heads
$\text{X}=6,7,8$
Here $\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
P(6)=Prob of getting 6 heads, 2 tails $=\text{ }^8\text{C}_6\big(\frac{1}{6}\big)^6\times\big(\frac{1}{2}\big)^2$
P(7) = Prob of getting 7 heads, 1 tails $=\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^7\times\big(\frac{1}{2}\big)^1$
P(8) = Prob of getting 8 heads, 0 tails $=\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8\times\big(\frac{1}{2}\big)^0$
The probability of getting at least 6 heads (not more than 2 tails) is then
$\text{ }^8\text{C}_6\big(\frac{1}{6}\big)^6\times\big(\frac{1}{2}\big)^2+\text{ }^8\text{C}_1\big(\frac{1}{2}\big)^7\times\big(\frac{1}{2}\big)^1+\text{ }^8\text{C}_2\big(\frac{1}{2}\big)^8\times\big(\frac{1}{2}\big)^0$
$=\frac{1}{256}+8\frac{1}{256}+28\frac{1}{256}=\frac{37}{256}$
View full question & answer→Question 2045 Marks
In a binomial distribution the sum and product of the mean and the variance are $\frac{25}{3}$ and $\frac{50}{3}$ respectively. Find the distribution.
AnswerLet n and p be the parameter of distribution binomial distribution. So
$\text{q}=1-\text{p}$ as $\text{p + q}=1$
$\text{ Mean + variance}=\frac{25}{3}$
$\text{np + npq}=\frac{25}{3}$
$\text{np(1+ q)}=\frac{25}{3}$
$\text{np}=\frac{25}{3(1+\text{ q})}\dots(1)$
$\text{Mean}\times\text{Variance}=\frac{50}{3}$
$\text{np}\times\text{npq}=\frac{50}{3}$
$\text{n}^2\text{p}^2\text{q}=-\frac{50}{3}$
$\Big[\frac{25}{3(1+\text{q})}\Big].\text{q}=\frac{50}{3}$ [using(1)]
$625\text{q}=\frac{50}{3}\big[9(1+\text{q})^2\big]$
$625\text{q}=150(1+\text{q})^2$
$25\text{q}=6(1+\text{q})^2$
$6+6\text{q}^2+12\text{q}-25\text{q}=0$
$6\text{q}^2-13\text{q}+6=0$
$6\text{q}^2-9\text{q}-4\text{q}+6=0$
$3\text{q}(2\text{q}-3)-2(2\text{q}-3)=0$
$(2\text{q}-3)(3\text{q}-2)=0$
$\Rightarrow2\text{q}-3=0$ or $3\text{q}-2=0$
$\Rightarrow\text{q}=\frac{3}{2}$ or $\text{q}=\frac{2}{3}$
Since $\text{q}\leq1,$ So
$\text{q}=\frac{2}{3}$
$\text{p}=1-\text{q}$
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
View full question & answer→Question 2055 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.
AnswerLet 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 $=\text{P}\Big(\frac{\text{C}}{\text{E}}\Big)=\frac{\text{P}(\text{C})\text{P}\Big(\frac{\text{E}}{\text{C}}\Big)}{\text{P}(\text{A})\text{P}\Big(\frac{\text{E}}{\text{A}}\Big)+\text{P}(\text{B})\text{P}\Big(\frac{\text{E}}{\text{B}}\Big)+\text{P}(\text{C})\text{P}\Big(\frac{\text{E}}{\text{C}}\Big)}$
$=\frac{\frac{20}{60}\times0.2}{\frac{10}{60}\times0.002+\frac{30}{60}\times0.02+\frac{20}{60}\times0.2}$
$=\frac{4}{4.62}=\frac{400}{462}=\frac{200}{231}$
View full question & answer→Question 2065 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?
AnswerLet $E_1, E_2$ 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
$=\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\times\text{P}(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}$
$=\frac{\frac{1}{3}}{\frac{1}{3}\Big(1+\frac{3}{4}+\frac{1}{2}\Big)}$
$=\frac{1}{\frac{9}{4}}$
$=\frac{4}{9}$
View full question & answer→Question 2075 Marks
Two dice are tossed. Find whether the following two events A and B are
independent:
$\text{A}=\left\{(\text{X},\text{Y}):\text{x}+\text{y}=11\right\}$ and $\text{B}=\left\{(\text{x,y}):\text{x}\neq5\right\}$
where (x, y) denotes a typical sample point.
AnswerWe have,
$\text{A}=\left\{(\text{X},\text{Y}):\text{x}+\text{y}=11\right\}$
And $\text{B}=\left\{(\text{x,y}):\text{x}\neq5\right\}$
$\therefore$ A = {(5, 6), (6, 5)}
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5) (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5) (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5) (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\Rightarrow\text{n}(\text{A})=2,\text{n}(\text{B})=30$ and $\text{n}(\text{A}\cap\text{B})=1$
$\therefore\text{P}(\text{A})=\frac{2}{36}=\frac{1}{18}$ and $\text{P}(\text{B})=\frac{30}{36}=\frac{5}{6}$
$\Rightarrow\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{5}{108}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{18}\neq\text{P}(\text{A})\cdot(\text{B})$
So, A and B are not independent.
View full question & answer→Question 2085 Marks
Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.
AnswerA, four balls are to be drawn without replacement and X denote the number of red balls drawn.
So, X is a random variable that can take values 0, 1, 2, 3 or 4.
Now,
P(X = 0) = P(All white balls) = P(WWWW) $=\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{1}{9}=\frac{1}{495},$
P(X = 1) = P(One of red balls and three white balls) = P(RWWW) + P(WRWW) + P(WWRW) + P(WWWR)
$=\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{8}{9}$
$=4\times\frac{8}{495}=\frac{32}{495}.$
P(X = 2) = P(Two red balls and two white balls) = P(RRWW) + P(RWRW) + P(RWWR) + P(WWRR) + P(W
P(X = 3) = P(Three red balls and one white ball) = P(RRRW) + P(RRWR) + P(RWRR) + P(WRRR)
$=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{4}{9}+\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{7}{10}\times\frac{6}{9}$
$=4\times\frac{56}{495}=\frac{224 }{495}.$
P(X = 4) = P(All red balls) = P(RRRR) $=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}$
$=\frac{70}{495}$
So, the probability distribution of X is as follows:
|
$\text{X}$
|
$0$
|
$1$
|
$2$
|
$3$
|
$4$
|
|
$\text{P}(\text{X})$
|
$\frac{1}{495}$
|
$\frac{32}{495}$
|
$\frac{168}{495}$
|
$\frac{224}{495}$
|
$\frac{70}{495}$
|
View full question & answer→Question 2095 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.
AnswerConsider 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 $=\frac{3}{7}$
View full question & answer→Question 2105 Marks
A letter is known to have come either from $\text{LONDON}$ or $\text{CLIFTON}$. On the envelope just two consecutive letters $\text{ON}$ are visible. What is the probability that the letter has come from $\text{CLIFTON}$?
Answersolution Consider events $\mathrm{E}_1, \mathrm{E}_2$ and $A$ events As:
$E_1=$ Letters come from $\ce{LONDON}$
$\mathrm{E}_2=$ Letters come from $\ce{CLIFTON}$
$E_3=$ Two consecutive letters visible on the envelope are on
$\mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{2}$
$\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2}$
$[$Since letters came either from $\ce{LONDON}$ or $\ce{CLIFTON]}$
$P\left(A \mid E_1\right)=P($ Two consecutive letters $\ce{ON}$ from $\ce{LONDON})$
$=\frac{2}{5}$
$[$Since $\ce{LONDON}$ has $2 - \ce{ON}$ and $5$ letters we consider one $'ON\ '$ as one letter$]$
$\mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}_2}\right)=\mathrm{P}($ Two consecutive letters $\ce{ON}$ from $\text{CLIFTON})$
$ =\frac{1}{6}$
$[$Since $\ce{CLIFTON}$ has one $'ON\ '$ nad $6$ letters considering $\ce{ON}$ as one letter$]$
$\mathrm{P}\left(\frac{\mathrm{E}_2}{\mathrm{~A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}_2}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}_2}\right)}$
$ =\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 $=\frac{5}{17}$
View full question & answer→Question 2115 Marks
Write the values of 'a' for which the following distribution of probabilities becomes a probability distributioin:
| $\text{X}=\text{x}_\text{i}:$ |
$-2$ |
$-1$ |
$0$ |
$1$ |
| $\text{P}(\text{X}=\text{x}_\text{i}):$ |
$\frac{1-\text{a}}{4}$ |
$\frac{1+2\text{a}}{4}$ |
$\frac{1-2\text{a}}{4}$ |
$\frac{1+\text{a}}{4}$ |
AnswerHere,
| $\text{X}:$ |
$-2$ |
$-1$ |
$0$ |
$1$ |
| $\text{P}(\text{X}):$ |
$\frac{1-\text{a}}{4}$ |
$\frac{1+2\text{a}}{4}$ |
$\frac{1-2\text{a}}{4}$ |
$\frac{1+\text{a}}{4}$ |
Now, $\sum\text{P}(\text{X})=\text{P}(-)+\text{P}(-1)+\text{P}(0)+\text{P}(1)$
$=\frac{1-\text{a}}{4}+\frac{1+2\text{a}}{4}+\frac{1-2\text{a}}{4}+\frac{1+\text{a}}{4}$
$=\frac{1-\text{a}-1+2\text{a}+1-2\text{a}+1+\text{a}}{4}$
$=1$
So, the sum of the probability must be positive and less than or equal to 1.
Now,
$0\leq\frac{1-\text{a}}{4}\leq1\Rightarrow0\leq1-\text{a}\leq4\Rightarrow-1\leq-\text{a}\leq3\Rightarrow1\geq\text{a}\geq-3$
$0\leq\frac{1+2\text{a}}{4}\leq1\Rightarrow0\leq1+2\text{a}\leq4\Rightarrow-1\leq2\text{a}\leq3\Rightarrow-\frac{1}{2}\leq\text{a}\leq\frac{3}{2}$
$0\leq\frac{1+\text{a}}{4}\leq1\Rightarrow0\leq1+\text{a}\leq4\Rightarrow-1\leq\text{a}\leq3$
Therefore, $-\frac{1}{2}\leq\text{a}\leq\frac{1}{2}$ View full question & answer→Question 2125 Marks
The probability distribution of a discrete random variable X is given as under:
| $\text{X}$ |
$1$ |
$2$ |
$4$ |
$2\text{A}$ |
$3\text{A}$ |
$5\text{A}$ |
| $\text{P}(\text{X})$ |
$\frac{1}{2}$ |
$\frac{1}{5}$ |
$\frac{3}{25}$ |
$\frac{1}{10}$ |
$\frac{1}{25}$ |
$\frac{1}{25}$ |
Calculate:
- The value of A if E(X) = 2.94
- Variance of X.
AnswerWe have
$\sum\text{X}\text{P}(\text{X})=\frac{1}{2}+\frac{2}{5}+\frac{2}{25}+\frac{2\text{A}}{10}+\frac{3\text{A}}{25}+\frac{5\text{A}}{25}$
$=\frac{25+20+24+10\text{A}+6\text{A}+10\text{A}}{50}$
$=\frac{69+26\text{A}}{50}$
Since, $\text{E}(\text{X})=\sum\text{X}\text{P}(\text{X})$
$\Rightarrow2.94=\frac{69+26\text{A}}{50}$
$\Rightarrow26\text{A}=50\times2.94-69$
$\Rightarrow\text{A}=\frac{147-69}{26}$
$=\frac{78}{26}=3$
We know that,
$\text{Var}(\text{X})=\text{E}(\text{X}^2)-[\text{E}(\text{X})]^2$
$=\sum\text{X}^2\text{P}(\text{X})-[\text{E}(\text{X})]^2$
$=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{4\text{A}^2}{10}+\frac{9\text{A}^2}{25}+\frac{25\text{A}^2}{25}-[\text{E}(\text{X})]^2$
$=\frac{25+40+96+20\text{A}^2+18\text{A}^2+50\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\text{A}^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{161+88\times(3)^2}{50}-[\text{E}(\text{X})]^2$
$=\frac{953}{50}-[2.94]^2$ $[\because\text{E}(\text{X})=2.94]$
$=19.06-8.6436$
$=10.4164$
View full question & answer→Question 2135 Marks
Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of kings.
AnswerLet X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
= P(no kings)
$=\frac{48}{52}\times\frac{48}{52}$
$=\frac{12\times12}{13\times13}$
$=\frac{144}{169}$
P(X = 1)
= P(1 king)
$=\frac{4}{52}\times\frac{48}{52}$
$=\frac{2\times12}{13\times13}$
$=\frac{24}{169}$
P(X = 2)
= P(2 kings)
$=\frac{4}{52}\times\frac{4}{52}$
$=\frac{1\times1}{13\times13}$
$=\frac{1}{169}$
Thus, the probability distribution of X is given by
|
$\text{X}$
|
$0$
|
$1$
|
$2$
|
|
$\text{P}(\text{X})$
|
$\frac{144}{169}$
|
$\frac{24}{169}$
|
$\frac{1}{169}$
|
View full question & answer→Question 2145 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.
AnswerA 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.
View full question & answer→Question 2155 Marks
The probability distribution of a random variable X is given below:
| $\text{X}$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{X})$ |
$\text{k}$ |
$\frac{\text{k}}{2}$ |
$\frac{\text{k}}{4}$ |
$\frac{\text{k}}{8}$ |
- Determine the value of k.
- Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}\geq2)$
- Find $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$
AnswerWe have,
| $\text{X}$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{X})$ |
$\text{k}$ |
$\frac{\text{k}}{2}$ |
$\frac{\text{k}}{4}$ |
$\frac{\text{k}}{8}$ |
- Since, $\sum_\limits{\text{i}=1}^\text{n}\text{P}_{\text{i}}=1,\text{i}=1,2, ....,\text{n}$ and $\text{P}_{\text{i}}\geq0$
$\therefore\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$\Rightarrow8\text{k}+4\text{k}+2\text{k}+\text{k}=8$
$\therefore\text{k}=\frac{8}{15}$
- $\text{P}(\text{X}\leq2)=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{(4\text{k}+2\text{k}+\text{k})}{4}=\frac{7\text{k}}{4}$
$=\frac{7}{4}\cdot\frac{8}{15}=\frac{14}{15}$
And $\text{P}(\text{X}\geq2)=\text{P}(3)=\frac{\text{k}}{8}$
$=\frac{1}{8}\cdot\frac{8}{15}=\frac{1}{15}$
- $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$
$=\frac{14}{15}+\frac{1}{15}=1$ View full question & answer→Question 2165 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.
AnswerAs, 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) $=\frac{5}{9}$
$\Rightarrow\ \text{P(Only A)}+\text{P(Only B)}=\frac{5}{9}$
$\Rightarrow\ \big[\text{P(A)}-\text{P}(\text{A}\cup\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cup\text{B})\big]=\frac{5}{9}$
$\Rightarrow\ \big[\text{P(A)}-\text{P}(\text{A})+\text{P}(\text{B})\big]\times\big[\text{P(B)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]=\frac{5}{9}$
$\Rightarrow\ \text{P(A)}\times\big[1-\text{P}(\text{B})\big]+\text{P(B)}\big[1-\times\text{P}(\text{A})\big]=\frac{5}{9}$
$\Rightarrow\ \text{p}\big[1-2\text{p}\big]+2\text{p}\big[1-\text{p}\big]=\frac{5}{9}$
$=\text{p}-2\text{p}^2+2\text{p}-2\text{p}^2=\frac{5}{9}$
$\Rightarrow\ 3\text{p}-4\text{p}^2=\frac{5}{9}$
$\Rightarrow\ 27\text{p}-36\text{p}^2=5$
$\Rightarrow\ 36\text{p}^2-27\text{p}+5=0$
View full question & answer→Question 2175 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 $3$? AnswerUrn $I$ 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 $=\frac{1}{40}$
View full question & answer→MCQ 2185 Marks
Let $E_1$ and $E_2 $ be two independent events such that $P(E_1) = P_1$ and $P(E_2) = P_2$. Describe in words of the events whose probabilities are:
- A
$P_1P_2$
- B
$(1 - P_1)P_2$
- C
$1 - (1 - P_1)(1 - P_2)$
- ✓
$P_1 + P_2- 2P_1P_2$
AnswerCorrect option: D. $P_1 + P_2- 2P_1P_2$
$\text{P}(\text{E}_1)=\text{P}_1$ and $\text{P}(\text{E}_2)=\text{P}_2$
$\text{P}_1\text{P}_2\Rightarrow\text{P}(\text{E}'_1)\cdot\text{P}(\text{E}_2)=\text{P}(\text{E}'_1\cap\text{E}_2)$
So $, E_1$ and $E_2$ occurs.
$(1-\text{P}_1)\text{P}_2=\text{P}(\text{E}'_1)\cdot\text{P}(\text{E}_2)=\text{P}(\text{E}'_1\cap\text{E}_2)$
So $, E_1$ does not occur but $E_2$ occurs.
$1-(1-\text{P}_1)(1-\text{P}_2)$
$=1-\text{P}(\text{E}_1)'\cdot\text{P}(\text{E}_2)'$
$=\text{P}(\text{E}'_1\cap\text{E}'_2)$
$=1-\big[1-\text{P}(\text{E}_1\cup\text{E}_2)\big]$
$=\text{P}(\text{E}_1\cup\text{E}_2)$
So, either $E_1$ or $E_2$ or both $E_1$ and $E_2$ occurs.
$\text{P}_1+\text{P}_2-2\text{P}_1\text{P}_2$
$=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)-2\text{P}(\text{E}_1)\cdot\text{P}(\text{E}_2)$
$=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)-2\text{P}(\text{E}_1-\text{E}_2)$
$=\text{P}(\text{E}_1\cup\text{E}_2)-\text{P}(\text{E}_1\cup\text{E}_2)$
So, either $E_1$ or $E_2$ occurs but not both.
View full question & answer→Question 2195 Marks
Suppose that $6\%$ of the people with blood group $O$ are left handed and $10\%$ of those with other blood groups are left handed $30\%$ of the people have blood group $O$. If a left handed person is selected at random, what is the probability that he/she will have blood group $O$?
AnswerGiven conditions can be represented as below:
| |
|
Blood group $ 'O'$
|
Other than blood group $'O'$
|
| $i$. |
Number of people
|
$30\%$ |
$70\%$ |
| $ii$. |
Percentage of left handed people
|
$6\%$ |
$10\%$ |
Let $E_1 =$ Event that the person selected is of blood group $O$
$E_2 =$ Event that the person selected is of other than blood group $O$
And $(E_3) =$ Event that selected person is left handed
$\therefore\text{P}(\text{E}_1)=0.30,\text{P}(\text{E}_2)=0.70$
$\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)=0.60$ and $\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)=0..10$
Using Baye’s theorem, we have
$\text{P}\Big(\frac{\text{E}_1}{\text{E}_3}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)\cdot\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)}$
$=\frac{0.30\times0.06}{0.30\times0.06+0.70\times0.10}$
$=\frac{0.0180}{0.0180+0.0700}$
$=\frac{0.0180}{0.0880}=\frac{180}{880}$
$=\frac{9}{44}$ View full question & answer→Question 2205 Marks
A and B are two events such that $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4}.$ Find:
- $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
- $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
- $\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)$
- $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)$
AnswerHere, $\text{P}(\text{A})=\frac{1}{2},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
- $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}$
- $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
- $\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cup\text{B})}{\text{P}(\text{B})}$ $=\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{3}}=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$
- $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B})}=\frac{1-\text{P}(\text{A}\cup\text{B})}{1-\text{P}(\text{B})}$
$=\frac{1-\big[\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})\big]}{1-\text{P}(\text{B})}$
$=\frac{1-\big[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\big]}{1-\frac{1}{3}}=\frac{1-\Big(\frac{5}{6}-\frac{1}{4}\Big)}{\frac{2}{3}}$
$=\frac{1-\frac{14}{24}}{\frac{2}{3}}=\frac{\frac{10}{24}}{\frac{2}{3}}$
$=\frac{30}{48}=\frac{5}{8}$ View full question & answer→Question 2215 Marks
$A$ discrete random variable $X$ has the probability distribution given as below:
| $X$ |
$0.5$ |
$1$ |
$1.5$ |
$2$ |
| $P(X)$ |
$k$ |
$k^2$ |
$2k^2$ |
$k$ |
- Find the value of $k$.
- Determine the mean of the distribution.
AnswerWe have,
| $X$ |
$0.5$ |
$1$ |
$1.5$ |
$2$ |
| $P(X)$ |
$k$ |
$k^2$ |
$2k^2$ |
$k$ |
- We know that $\sum\limits_\text{i=1}^{\text{n}}\text{P}_{\text{i}}=1,$ where $\text{P}_{\text{i}}\geq0$
- $\Rightarrow P_1+ P_2 + P_3 + P_4= 1$
$\Rightarrow k + k^2 + 2k^2 + k = 1$
$\Rightarrow 3k^2 + 2k - 1 = 0$
$\Rightarrow 3k^2 + 3k - k - 1 = 0$
$\Rightarrow 3k(k + 1) - 1(k + 1) = 0$
$\Rightarrow (3k - 1)(k + 1) = 0$
$\Rightarrow\text{k}=\frac{1}{3}$ and $k = -1$
Since, $\text{K}\geq0,$ we take $\text{k}=\frac{1}{3}$
- Mean of the distribution $(\mu)=\text{E}(\text{X})=\sum\limits_{\text{i=1}}^{\text{n}}\text{x}_{\text{i}}\text{P}_{\text{i}}$
- $=0.5(\text{k})+1(\text{k}^2)+1.5(2\text{k}^2)+2(\text{k})$
$=4\text{k}^2+2.5\text{k}$
$=4\cdot\frac{1}{9}+2.5\cdot\frac{1}{3}\Big[\because\text{k}=\frac{1}{3}\Big]$
$=\frac{4+7.5}{9}=\frac{23}{18}$
View full question & answer→Question 2225 Marks
The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P (X = 1) and P (X ≥ 2).
AnswerGiven that, parameter for binomial distribution are n and p.
Also, $\text{Mean = np}=16\dots(1)$
$\text{Variance = npq = 8}\dots(2)$
Dividing (2) ane (1)
$\text{q}=\frac{1}{2}$
So, $\text{p}=1-\frac{1}{2}$ [as p + q = 1]
$\text{p}=\frac{1}{2}$
put the value of p in equation (1),
$\text{np}=16$
$\text{n}\big(\frac{1}{2}\big)=16$
$\text{n}=32$
Hence, binomial distribution is given by,
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{32-\text{r}}\dots(3)$
$\text{P(X}=0)$
$=\text{ }^{32}\text{c}_0\big(\frac{1}{2}\big)^0\big(\frac{1}{2}\big)^{32-0}$ [Using (3)]
$=\big(\frac{1}{2}\big)^{32}$
$=\text{P(X}=1)$
$=\text{ }^{32}\text{c}_1\big(\frac{1}{2}\big)^1\big(\frac{1}{2}\big)^{32-1}$
$=32.\frac{1}{2}\big(\frac{1}{2}\big)^{31}$
$=\big(\frac{1}{2}\big)^{27}$
$\text{P(X}\geq2)$
$=1-\big[\text{P(X}=0)+\text{P(X}=1)\big]$
$=1-\Big[\big(\frac{1}{2}\big)^{32}+\big(\frac{1}{2}\big)^{27}\Big]$
$=1-\big(\frac{1}{2}\big)^{27}\big(\frac{1}{32}+1\big)$
$=1-\big(\frac{1}{2}\big)^{27}\big(\frac{33}{32}\big)$
$=1-\frac{33}{2^{32}}$
Hence
$\text{P(X}=0)=\big(\frac{1}{2}\big)^{32},\text{P(X = 1})=\big(\frac{1}{2}\big)^{27},\text{P(X}\geq2)=1-\frac{33}{2^{32}}$
View full question & answer→Question 2235 Marks
If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).
AnswerLet n and p the parameter of binomial distribution.
Given,
$\text{Mean = np}=4\dots(1)$
$\text{Variance = npq}=2\dots(2)$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$
$\text{p}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distributon is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}\dots(3)$
$\text{P(X}\geq5)$
$=\text{P(X}=5)+\text{P(X}=6)+\text{P(X}=7)+\text{P(X}=8)$
$=\text{ }^{8}\text{C}_5\big(\frac{1}{2}\big)^{5}\big(\frac{1}{2}\big)^{3}+\text{ }^{8}\text{C}_6\big(\frac{1}{2}\big)^{6}\big(\frac{1}{2}\big)^{2}+\text{ }^{8}\text{C}_7\big(\frac{1}{2}\big)^{7}\big(\frac{1}{2}\big)+\text{ }^{8}\text{C}_8\big(\frac{1}{2}\big)^{8}$
[Using equation (3)]
$=\frac{8\times7\times6}{3\times2}\big(\frac{1}{2}\big)^8+\frac{8\times7}{2}\big(\frac{1}{2}\big)^8+8\big(\frac{1}{2}\big)^8+\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^8\big[56+28+8+1\big]$
$=\frac{93}{256}$
$\text{P(X}\geq5)=\frac{93}{256}$
View full question & answer→Question 2245 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.
AnswerUrn $I$ 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 $1$ 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 $=\frac{2}{9}$
View full question & answer→Question 2255 Marks
Three bags contain a number of red and white balls as follows: Bag $1 : 3$ red balls, Bag $2 : 2$ red balls and $1$ white ball Bag $3 : 3$ white balls.
The probability that bag i will be chosen and $a$ ball is selected from it is $\frac{\text{i}}{6}, i = 1, 2, 3. $ What is the probability that:
- $A$ red ball will be selected?
- $A $ white ball is selected?
AnswerWe have, Bag $I: 3$ red balls and $0$ white bail. Bag $II : 2$ red balls and $1$ white ball. Beg $III: 0$ red ball and $3$ white balls
- Let $E_1, E_2$ and $E_3$ be the events that bag $I,$ bag $II$ and bag $III$ is selected and a ball is chosen from it.
$\text{P}(\text{E}_1)=\frac{1}{6},\text{P}(\text{E}_2)=\frac{2}{6}$ and $\text{P}(\text{E}_3)=\frac{3}{6}$
Let $E$ be the event that $a$ red ball is selected.
Then, probability that $a$ red ball will be selected is,
$\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big )+\text{P}(\text{E}_2)\cdot\Big(\frac{\text{E}}{\text{E}_2}\Big )+\text{P}(\text{E}_3)\cdot\Big(\frac{\text{E}}{\text{E}_3}\Big )$
$=\frac{1}{6}\cdot\frac{3}{3}+\frac{2}{6}\cdot\frac{2}{3}+\frac{3}{6}\cdot0$
$=\frac{1}{6}+\frac{2}{9}+0$
$=\frac{3+4}{18}=\frac{7}{18}$
- Let $E\ '$ be the event that $a$ white ball is selected. Then
$\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big )+\text{P}(\text{E}_2)\cdot\Big(\frac{\text{E}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\cdot\Big(\frac{\text{E}}{\text{E}_3}\Big)$
$=\frac{1}{6}\cdot0+\frac{2}{6}\cdot\frac{1}{3}+\frac{3}{6}\cdot1$
$=\frac{1}{9}+\frac{3}{6}=\frac{11}{18}$ View full question & answer→Question 2265 Marks
Two dice are thrown together and the total score is noted. The events 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.
AnswerTwo dice are thrown together i.e.,
$\therefore$ n(S) = 36, where S is the sample space.
Event 'E' is 'a total of 4'
$\therefore$ E = {(2, 2), (3, 1), (1.3)}
Event 'F' is 'a total of 9 or more'
$\therefore$ F = {(3, 6), (6, 3), (4, 5), (4, 6), (5, 4), (6, 4), (5, 5), (5, 6), (6 5), (6, 6)}
Event 'G' is 'a total divisible by 5'
$\therefore$ G = {(1, 4), (4, 1), (2, 3). (3, 2), (4, 6), (6, 4), (5 5)}
Here, $(\text{E}\cap\text{F})=\phi$ and $(\text{E}\cap\text{G})=\phi$
Also, $(\text{F}\cap\text{G})=\left\{(4,6),(6,4),(5,5)\right\}$
$\therefore\text{P}(\text{E})=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{3}{36}=\frac{1}{12}$
$\text{P}(\text{F})=\frac{\text{n}(\text{F})}{\text{n}(\text{S})}=\frac{10}{36}=\frac{5}{18}$
$\text{P}(\text{G})=\frac{\text{n}(\text{G})}{\text{n}(\text{S})}=\frac{7}{36}$
$\text{P}(\text{F}\cap\text{G})=\frac{3}{36}=\frac{1}{12}$
and $\text{P}(\text{F})\cdot\text{P}(\text{G})=\frac{5}{18}\cdot\frac{7}{36}=\frac{35}{648}$
So, $\text{P}(\text{F}\cap\text{G})\neq\text{P}(\text{F})\cdot\text{P}(\text{G})$
Hence, there is no pair which is independent.
View full question & answer→Question 2275 Marks
Assume that the probability that a bomb dropped from an aeroplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability that.
- exactly 2 will strike the target.
- at least 2 will strike the target.
Answerprobabilty that bomb strikes a target p = 0.2
Probability that a bomb misses the target = 0.8
n = 6
let X = number of bombs that strike the target
P(X = 2) = exactly 2 bombs strike the target
$=\text{ }^6\text{C}_2\big(\frac{2}{10}\big)^2\times\big(\frac{8}{10}\big)^4=15\times\frac{16384}{10^6}=0.24576$
$\text{P(X}\geq2)=$ at least 2 bombs strike the target
$=1-\text{P(X}<2)$
$=1-\big[\text{P(X}=0)+\text{P(X=1})\big]$
$=1-\big[\text{ }^6\text{C}_0\big(\frac{2}{10}\big)^0\times\big(\frac{8}{10}\big)^6+\text{ }^6\text{C}_1\big(\frac{2}{10}\big)^1\times\big(\frac{8}{10}\big)^6\big]$
$=1-\big[0.0.262144+0.393216\big]=1-0.65536$
$=0.34464$
View full question & answer→Question 2285 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,
- A and B.
- B and C.
- C and A.
AnswerA 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} $\text{P(C)}=\frac{2}{8}$ $\text{P(C)}=\frac{1}{4}$ $\text{A}\cap\text{B}\{\text{HHH, HHT}\}$ $\text{P}(\text{A}\cap\text{B})=\frac{2}{8}$ $=\frac{1}{4}$ $\text{B}\cap\text{C}=\{\text{HHT, THH}\}$ $\text{P}(\text{B}\cap\text{C})=\frac{2}{8}$ $\text{P}(\text{B}\cap\text{C})=\frac{1}{4}$ $(\text{A}\cap\text{C})=\{\text{HHT}\}$ $\text{P}(\text{A}\cap\text{C})=\frac{1}{8}$
- $\text{P(A)} \text{ P(B)}=\frac{1}{2}\times\frac{1}{2}$
$=\frac{1}{4}$
$\text{P(A) }\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
Hence, A and B are independent events.
- $\text{P(B) }\text{P(C)}=\frac{1}{2}\times\frac{1}{4}$
$=\frac{1}{8}$
$\text{P(B) }\text{P(C)}\neq\text{P}(\text{B}\cap\text{C})$
So, B and C are not independent events.
- $\text{P(A) }\text{P(C)}=\frac{1}{2}\times\frac{1}{4}$
$=\frac{1}{8}$
$\text{P(A) }\text{P(C)}=\text{P}(\text{A}\cap\text{C})$
Hence, A and C are independent events. View full question & answer→Question 2295 Marks
From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
AnswerLet getting a defective bulb in a trial be a success.
We have,
p = probability of getting a defective bulb $=\frac{5}{15}=\frac{1}{3}$ and
q = probability of getting non - defective bulb $=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Let X denote the number of success in a sample of 4 trials. Then,
X follows binomial distribution with parameter $\text{n}=4$ and $\text{p}=\frac{1}{3}$
$\therefore\text{P(X = r})=\text{ }^4\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(4-\text{r})}=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{(4-\text{r})}=\frac{\text{ }^4\text{c}_{\text{r}}2^{(4-\text{r})}}{3^4},$ where $\text{r}=0,1,2,3,4$
i.e.
$\text{P(X}=0)=\frac{\text{ }^4\text{C}_02^4}{3^4}=\frac{16}{81},$
$\text{P(X}=1)=\frac{\text{ }^4\text{C}_12^3}{3^4}=\frac{32}{81},$
$\text{P(X}=2)=\frac{\text{ }^4\text{C}_22^2}{3^4}=\frac{24}{81},$
$\text{P(X}=3)=\frac{\text{ }^4\text{C}_32^1}{3^4}=\frac{8}{81},$
$\text{P(X}=4)=\frac{\text{ }^4\text{C}_42^0}{3^4}=\frac{1}{81}$
So, the probability distribution of X is given as follows:
| $\text{X}:$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $\text{P(X):}$ |
$\frac{16}{81}$ |
$\frac{32}{81}$ |
$\frac{24}{81}$ |
$\frac{8}{81}$ |
$\frac{1}{81}$ |
Now,
$\text{Mean, E(X)}=0\times\frac{16}{81}+1\times\frac{32}{81}+2\times\frac{24}{81}+3\times\frac{8}{81}+4\times\frac{1}{81}$
$=\frac{32+48+24+4}{81}$
$=\frac{108}{81}$
$=\frac{4}{3}$
Note: we can also calculate the mean of the binomial distribution by
$\text{Mean, E(X) = np}=4\times\frac{1}{3}=\frac{4}{3}$ View full question & answer→Question 2305 Marks
Two biased dice are thrown together. For the first die $\text{P}(6)=\frac{1}{2},$ the other scores being equally likely while for the second die, $\text{P}(1)=\frac{2}{5}$ and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.
AnswerFor first die, $\text{P}(6)=\frac{1}{2}$ and $\text{P}(\text{not}6)=\frac{1}{2}$
$\Rightarrow\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)+\text{P}(5)=\frac{1}{2}$
$\Rightarrow\text{P}(1)=\frac{1}{10}$ and $\text{P}(\text{not}1)=\frac{9}{10}$ $\big[\because\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)+\text{P}(5)\big]$
For second die, $\text{P}(1)=\frac{2}{5}$ and $\text{P}(\text{not}1)=1-\frac{2}{5}=\frac{3}{5}$
Let X = Number of one's seen
P(X = 0) = P(not 1 on first die) × P(not 1 on second die)
$=\frac{9}{10}\cdot\frac{3}{5}=\frac{27}{50}=0.54$
P(X = 1) = P(1 on first die) × P(not 1 on second die) + P(1 on second die) × P(not 1 on first die)
$=\frac{1}{10}\cdot\frac{3}{5}+\frac{9}{10}\cdot\frac{2}{5}=\frac{21}{50}=0.42$
P(X = 2) = P(not 1 on first die) × P(not 1 on second die)
$=\frac{1}{10}\cdot\frac{2}{5}=0.04$
Hence, the required probability distribution is as below.
|
X
|
0
|
1
|
2
|
|
P(X)
|
0.54
|
0.42
|
0.04
|
View full question & answer→Question 2315 Marks
It is known that 60% of mice inoculated with a serum are protected from a certain disease. If 5 mice are inoculated, find the probability that.
- none contract the disease.
- more than 3 contract the disease.
AnswerLet X be the number of mice that contract the disease. Then, X follows a binomial distribution with n = 5. Let p be the probability of mice that cantract the disease. $\therefore\text{p}=0.4$ and $\text{q}=0.6$ Hence, the distribution is given by $\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}(0.4)^{\text{r}}(0.6)^{5-\text{r}},\text{r}=0,1,2,3,4,5$
- $\text{P(X}=0)=\text{ }^5\text{C}_0(0.4)^0(0.6)^{5-0}$
$=(0.6)^5$
$=0.0778$
- $\text{P(X}>3)=\text{P(X}=4)+\text{P(X}=5)$
$=\text{ }^5\text{C}_4(0.4)^4(0.6)^{5-4}+\text{ }^5\text{C}_5(0.4)^5(0.6)^{5-5}$
$=0.0768+0.01024$
$=0.08704$ View full question & answer→Question 2325 Marks
Two probability distributions of the discrete random variable $X$ and $Y$ are given below.
| $\text{X}$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{X})$ |
$\frac{1}{5}$ |
$\frac{2}{5}$ |
$\frac{1}{5}$ |
$\frac{1}{5}$ |
| $\text{Y}$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{Y})$ |
$\frac{1}{5}$ |
$\frac{3}{10}$ |
$\frac{2}{5}$ |
$\frac{1}{10}$ |
Prove that $E(Y^2) = 2E(X)$. AnswerSince, we have to prove that $, E(Y^2) = 2E(X) .......(i)$
Taking $\text{L.H.S}$. of equation $(i),$ we have
$\text{E}(\text{Y}^2)=\sum\text{Y}^2\text{P}(\text{Y})$
$=0\cdot\frac{1}{5}+1\cdot\frac{3}{10}+4\cdot\frac{2}{5}+9\cdot\frac{1}{10}$
$=\frac{3}{10}+\frac{8}{5}+\frac{9}{10}$
$=\frac{28}{10}=\frac{14}{5}$
$\Rightarrow\text{E}(\text{Y})^2=\frac{14}{5}\ ......(\text{ii})$
Now taking $\text{R.H.S}$. of equation $(i),$ we get
$\text{E}(\text{X})=\sum\text{XP}(\text{X})$
$=0.\frac{1}{5}+1\cdot\frac{2}{5}+2\cdot\frac{1}{5}+3\cdot\frac{1}{5}=\frac{7}{5}$
$2\text{E}(\text{X})=\frac{14}{5}\ ......(\text{iii})$
Thus, from equations $(ii)$ and $(iii),$ we get
$E(Y^2) = 2E(X)$
Hence proved.
View full question & answer→Question 2335 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 $B.$
AnswerBag $A$ 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 $=\frac{25}{52}$
View full question & answer→Question 2345 Marks
If a white ball is selected, what is the probability that it came from:
- Bag 2
- Bag 3
Answerwe have,
- $\text{P}\Big(\frac{\text{E}_2}{\text{F}}\Big)=\frac{\text{P}(\text{E}_2)\cdot\Big(\frac{\text{F}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_2}\Big )+\text{P}(\text{E}_3)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_3}\Big)}$
$=\frac{\frac{2}{6}\cdot\frac{1}{3}}{\frac{1}{6}\cdot0+\frac{2}{6}\cdot\frac{1}{3}+\frac{3}{6}\cdot1}=\frac{\frac{2}{18}}{\frac{2}{18}+\frac{3}{6}}$
$=\frac{\frac{2}{18}}{\frac{2+9}{18}}=\frac{2}{11}$
- $\text{P}\Big(\frac{\text{E}_3}{\text{F}}\Big)=\frac{\text{P}(\text{E}_3)\cdot\Big(\frac{\text{F}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_2}\Big )+\text{P}(\text{E}_3)\cdot\text{P}\Big(\frac{\text{F}}{\text{E}_3}\Big)}$
$=\frac{\frac{3}{6}\cdot1}{\frac{1}{6}\cdot0+\frac{2}{6}\cdot\frac{1}{3}+\frac{3}{6}\cdot1}$
$=\frac{\frac{3}{6}}{\frac{2}{18}+\frac{3}{6}}=\frac{\frac{3}{6}}{\frac{2+9}{18}}$
$=\frac{9}{11}$ View full question & answer→Question 2355 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 $X$?
AnswerConsider $E_1, E_2$ 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 $=\frac{1}{10}$
View full question & answer→Question 2365 Marks
If X is the number of tails in three tosses of a coin, determine the standard deviation of X.
AnswerGiven that, random variable X is the number of tails in three tosses of a coin.
So, X = 0, 1, 2, 3.
$\Rightarrow\text{P}(\text{X}=\text{x})={^{\text{n}}}\text{C}_{\text{x}}(\text{p})^{\text{x}}\text{q}^{\text{n}-\text{x}}$
Where $\text{n}=3,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$ and x = 0, 1, 2, 3
| $\text{X}$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{X})$ |
$\frac{1}{8}$ |
$$$\frac{3}{8}$ |
$\frac{3}{8}$ |
$\frac{1}{8}$ |
| $\text{XP}(\text{X})$ |
$0$ |
$\frac{3}{8}$ |
$\frac{3}{4}$ |
$\frac{3}{8}$ |
| $\text{X}^2\text{P}(\text{X})$ |
$0$ |
$\frac{3}{8}$ |
$\frac{3}{2}$ |
$\frac{9}{8}$ |
We know that, $\text{Var}(\text{X})=\text{E}(\text{X}^2)-\big[\text{E}(\text{X})\big]^2\ ......(\text{i})$
Where $\text{E}(\text{X}^2)=\sum_\limits{\text{i}=1}^\text{n}\text{x}^2_\text{i}\text{P}(\text{X}_\text{i})$ and $\text{E}(\text{X})=\sum_\limits{\text{i}=1}^\text{n}\text{x}_\text{i}\text{P}(\text{X}_\text{i})$
$\therefore\text{E}(\text{X}^2)=\sum_\limits{\text{i}=1}^\text{n}\text{x}^2_\text{i}\text{P}(\text{X}_\text{i})$ $=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}=3$
and $\big[\text{E}(\text{X})\big]^2=\bigg[\sum_\limits{\text{i}=1}^\text{n}\text{x}_\text{i}\text{P}(\text{X}_\text{i})\bigg]$ $=\bigg[0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}\bigg]^2=\bigg[\frac{12}{8}\bigg]^2=\frac{9}{4}$
$\therefore\text{Var}(\text{X})=3-\frac{9}{4}=\frac{3}{4}$
and standard deviation of X $=\sqrt{\text{Var}(\text{X})}=\sqrt\frac{3}{4}=\frac{\sqrt{3}}{2}$ View full question & answer→Question 2375 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.
AnswerGiven,
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}$
View full question & answer→Question 2385 Marks
Find the binomial distribution whose mean is 5 and variance $\frac{10}{3}.$
AnswerLet n and p be the parameters of binomial distribution.
Given, $\text{Mean = np}=5\dots(1)$
$\text{Variance = npq}=\frac{10}{3}\dots(2)$
Dividing (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{\frac{10}{3}}{5}$
$\text{q}=\frac{2}{3}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
Put the value of p in equation (1),
$\text{np}=5$
$\text{n}=5\times3$
$\text{n}=15$
Hence, the binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{15}\text{c}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{15-\text{r}}$
$\text{r}=0,1,2,\dots15$
View full question & answer→Question 2395 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$.
AnswerLet $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}$
View full question & answer→Question 2405 Marks
Let X denot the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
$\text{P}(\text{X = x})=\begin{cases}\text{kx},&\text{if}\text{ x}=0\text{ or }1\\2\text{kx},&\text{if x = 2}\\\text{k}(5-\text{x}),&\text{if x = 3 or 4}\\0,&\text{if x > 4}\end{cases}$
where k is a positive constant. Find the value of k. Also find the probability that you will get addmission in
- Exactly one college.
- At most two colleges.
- At least two colleges.
AnswerThe probability distribution of X is
| X |
0 |
1 |
2 |
3 |
4 |
| P(X) |
0 |
k |
4k |
2k |
k |
The given distribution is a probability distribution.
$\therefore\ \sum\text{p}_\text{i}=1$
⇒ 0 + k + 4k + 2k + k
⇒ 8k = 1
⇒ k = 0.125
- P(getting admission in exactly one college) $= \text{P}(\text{X }= 1) = \text{k} = 0.125$
- P(getting admission in at most 2 colleges) $=\text{P}(\text{X}\leq2)=0+\text{k}+4\text{k}=5\text{k}=0.625$
- P(getting admission in atleast 2 colleges) $=\text{P}(\text{X}\geq2)=4\text{k}+2\text{k}+\text{k}=7\text{k}=0.875$
View full question & answer→Question 2415 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$.
AnswerUrn $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}$
View full question & answer→Question 2425 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 $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 are 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?
AnswerLet $E_1 =$ Event that item is manufactured on $ A$.
$E_{2 }=$ Event that an item is manufactured on $B$.
$E_3=$ Event that an item is manufactured on $C$.
Let $E$ be the event that an item is defective.
$\therefore\text{P}(\text{E}_1)=\frac{50}{100}=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{30}{100}=\frac{3}{10}$
and $\text{P}(\text{E}_3)=\frac{20}{100}=\frac{1}{5}$
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{2}{100}=\frac{1}{50}$
$\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{2}{100}=\frac{1}{50}$
and $\text{P}\Big(\frac{\text{E}}{\text{E}_3}\Big)=\frac{3}{100}$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}$
$=\frac{\frac{1}{100}}{\frac{5+3+3}{500}}=\frac{5}{11}$
View full question & answer→Question 2435 Marks
The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}.$ Find the probability that among 5 components tested.
- exactly 2 will survive.
- at most 3 will survive.
AnswerLet p be the probability that componet survive the shock test.So $\text{p}=\frac{3}{4}$ $\text{q}=1-\frac{3}{4}$ [Since p + q = 1] $\text{q}=\frac{1}{4}$ Let X denote the random variable representing the number of components that survive out of n components is given by $\text{P(X = r } ) \ =\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{5-\text{r}}\dots(1)$
- Probability that exactly 2 will survive the shock test
$=\text{P(X}=2)$
$=\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}$
$=\frac{5.4}{2}\big(\frac{9}{16}\big)\big(\frac{1}{64}\big)$
$=\frac{45}{512}=0.0879$
Probability that exactly 2 survive $=0.0879$
- P( atmost 3 will survive) $=\text{P(X}\leq3)$
$=\text{P(X}=0)+\text{P(X}=1)+\text{P(X}=2)+\text{P(X}=3)$
$=\text{ }^5\text{C}_0\big(\frac{3}{4}\big)^0\big(\frac{1}{4}\big)^{5-0}+\text{ }^5\text{C}_1\big(\frac{3}{4}\big)^1\big(\frac{1}{4}\big)^{5-1}$
$+\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}+\text{ }^5\text{C}_3\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^{5-3}$
$=\big(\frac{1}{4}\big)^5+5\big(\frac{3}{4}\big)\big(\frac{1}{4}\big)^4+10\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^3+10\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^2$
$=\frac{1+15+90+270}{1024}$
$=\frac{376}{1024}$
$=0.3672$ View full question & answer→Question 2445 Marks
A bag contains $4$ white and $5$ black balls. Another bag contains $9$ white and $7$ black balls. $A$ ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.
AnswerHere, $W_1 = \{4$ white balls$\}$ and $B_1 = \{5$ black balls$\}$
And $W_2 = \{9$ white balls$\}$ and $B_2 = \{7$ black balls$\}$
Let $E_1$ is event that the ball transferred from the first bag is white and $E_2$ is the event that the ball transferred from the bag black.
Also, $E$ is the event that the bell drawn from the second bag is white.
$\therefore\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{10}{17}$
$\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{9}{17}$
And $\text{P}(\text{E}_1)=\frac{4}{9}$ and $\text{P}(\text{E}_2)=\frac{5}{9}$
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{4}{9}\cdot\frac{10}{17}+\frac{5}{9}\cdot\frac{9}{17}$
$=\frac{40+45}{153}=\frac{85}{153}=\frac{5}{9}$
View full question & answer→Question 2455 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?
AnswerLet $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.
View full question & answer→Question 2465 Marks
A letter is known to have come either from $\text{TATA NAGAR}$ or from $\text{CALCUTTA}$. On the envelope, just two consecutive letter $TA$ are visible. What is the probability that the letter came from $\text{TATA NAGAR}$.
AnswerLet $E_1$ be the event that a letter is from $\text{TATA NAGAR}$
$E_2$ be the event that letter is from $\text{CALCUTTA}$.
And $3 E$ be the event that two consecutive letters $TA$ are visible on envelope
$\therefore\text{P}(\text{E}_1)=\frac{1}{2}$ and $\text{P}(\text{E}_2)=\frac{1}{2}$
Also, if letter is from $\text{TATA NAGAR}$, we see that the events of two consecutive letters visible
are $\text{{TA, AT, TA, AN, NA, AG, GA, AR}}$
And if the letter is from $\text{CALCUTTA}$, we see that the events of getting two consecutive letters visible
are $\text{{CA, AL, LC, CU, UT, TT, TA}}$
Thus $\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)=\frac{2}{8}$ and $\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)=\frac{1}{7}$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}_3}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\cdot\frac{2}{8}}{\frac{1}{2}\cdot\frac{2}{8}+\frac{1}{2}\cdot\frac{1}{7}}$
$=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}=\frac{\frac{\frac{1}{8}}{22}}{8\times14}$
$=\frac{\frac{1}{8}}{\frac{11}{56}}=\frac{7}{11}$
View full question & answer→Question 2475 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).$
AnswerWe 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}$ 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}$
View full question & answer→Question 2485 Marks
There are three urns containing $2$ white and $3$ black balls, $3$ white and $2$ black balls, and $4$ white and $1$ black balls, respectively. There is an equal probability of each urn being chosen. $A$ ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
AnswerLet $U_{1 }= (2 $ white $, 3$ black balls$)$
$U_{2 }= (3 $ white $, 2$ black balls$)$
And $U_{3 }= (4$ white, 1 black balls$)$
$\therefore\text{P}(\text{U}_1)=\text{P}(\text{U}_2)=\text{P}(\text{U}_3)=\frac{1}{3}$
Let $E_1$ be the event that a ball is chosen from urn $U_1, E_{2 }$ be the event that $a$ ball is chosen from urn $U_{1 }$ and $E_{3 }$ be the event that a ball is chosen from urn $U_3$
Also, $\text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\text{P}(\text{E}_3)=\frac{1}{3}$
Now, let $E$ be the event that white ball is drawn.
$\therefore\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{2}{5},\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{2}{5},\text{P}\Big(\frac{\text{E}}{\text{E}_3}\Big)=\frac{4}{5}$
Now, $\text{P}\Big(\frac{\text{E}_2}{\text{E}}\Big)=\frac{\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)\text{P}(\text{E}_3)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{3}\cdot\frac{3}{5}}{\frac{1}{3}\cdot\frac{2}{5}+\frac{1}{3}\frac{3}{5}+\frac{1}{3}\cdot\frac{4}{5}}$
$=\frac{\frac{3}{15}}{\frac{2}{15}+\frac{3}{15}+\frac{4}{15}}$
$=\frac{3}{9}=\frac{1}{3}$
View full question & answer→Question 2495 Marks
Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X.$
Answer$S = (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2),(3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3),$
$ (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),$
$ (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6) n(S) = 36$
Let $A$ denotes the sum of the numbers
$= 2,$ B denotes the sum of the numbers
$= 3 C$ denotes the sum of the numbers
$= 4, D$ denotes the sum of the numbers
$= 5 E$ denotes the sum of the numbers
$= 6, F$ denotes the sum of the numbers
$= 7 G$ denotes the sum of the numbers
$= 8, H$ denotes the sum of the numbers
$= 9 I$ denotes the sum of the numbers
$= 10, J$ denotes the sum of the numbers
$= 11 K$ denotes the sum of the numbers
$= 12 A = {1, 1}, n(A) = 1,$
$P(A) =\frac{1}{36} B = {(1, 2), (2, 1)}, n(B) = 2, P(A) =\frac{2}{36} C = {(1, 3), (2, 2), (3, 1)}, n(C) $$= 3, P(A) =\frac{3}{36} D = {(1, 4), (2, 3), (3, 2), (4, 1)}, n(D) = 4, P(A) =\frac{4}{36} E $$= {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}, n(E) = 5, P(A) =\frac{5}{36} F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, n(F) $$= 6, P(A) =\frac{6}{36} G = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, n(G) = 5, P(A) =\frac{5}{36} H$$ = {(3, 6), (4, 5), (5, 4), (6, 3)}, n(H) = 4, P(A) =\frac{4}{36} I = {(4, 6), (5, 5), (6, 4)}, n(I) $$= 3, P(A) =\frac{3}{36} J = {(5, 6), (6, 5)}, n(J) = 2, P(A) =\frac{2}{36} K = {6, 6}, n(K) = 1, P(A) =\frac{1}{36}$
| $x_i$ |
$P(x_i)$ |
$x_i$ |
$P(x_i)$ |
$x_i$ |
$P(x_i)$ |
$2$
$3$
$4$
$5$ |
$\frac{1}{36}$
$\frac{2}{36}$
$\frac{3}{36}$
$\frac{4}{36}$ |
$6$
$7$
$8$
$9$ |
$\frac{5}{36}$
$\frac{6}{36}$
$\frac{5}{36}$
$\frac{4}{36}$ |
$10$
$11$
$12$
|
$\frac{3}{36}$
$\frac{2}{36}$
$\frac{1}{36}$
|
Mean $=\mu=\sum\text{p}_i\text{x}_i$
$=\frac{1}{36}\times2+\frac{2}{36}\times3+\frac{3}{36}\times4+\frac{4}{36}\times5+\frac{5}{36}\times6+\frac{6}{36}\times7+\frac{5}{36}\times8+\frac{4}{36}\times9+\frac{3}{36}\times10+\frac{2}{36}\times11+\frac{1}{36}\times12$
$=\frac{1}{36}(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) =\frac{252}{36}=7$
$\text{Now}\ \sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2=\frac{1}{36}(2)^2+\frac{2}{36}(3)^2+\frac{3}{36}(4)^2+\frac{4}{36}(5)^2+\frac{5}{36}(6)^2+\frac{6}{36}(7)^2+\frac{5}{36}(8)^2$$+\frac{4}{36}(9)^2+\frac{3}{36}(10)^2+\frac{2}{36}(11)^2+\frac{1}{36}(12)^2$
$=\frac{1}{36} (4 + 18 + 48 + 100 + 180 + 294 + 320 + 300 + 242 + 144) =\frac{1}{36}\times1974=\frac{329}{6}$ Variance $\sum\text{p}_i\text{x}^2_i\ -\Big(\sum\text{p}_i\text{x}_i\Big)^2=\frac{329}{6}-(7)^2=54.83-49=5.83$ Standard deviation $\sqrt{5.83}=2.4\ (\text{nearly})$ View full question & answer→Question 2505 Marks
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
- 5 successes?
- at least 5 successes?
- at most 5 successes?
AnswerThe repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials. Probability of getting an odd number in a single throw of a die is, $\text{p}=\frac{3}{6}=\frac{1}{2}$ $\therefore\ \text{q}=1-\text{p}=\frac{1}{2}$ X has a binomial distribution. Therefore, P(X = x) $=\ ^\text{n}\text{C}_{\text{n}-\text{x}}\text{q}^{\text{n-x}}\text{p}^\text{x},\ \text{where}\ \text{n}=0,\ 1,\ 2\ ...\text{n}$ $=\ ^6\text{C}_\text{x}\Bigg(\frac{1}{2}\Bigg)^{6-\text{x}}.\Bigg(\frac{1}{2}\Bigg)^\text{x}$ $=\ ^6\text{C}_\text{x}\Bigg(\frac{1}{2}\Bigg)^{6}$
- P(5 successes) = P(X = 5)
$=\ ^6\text{C}_\text{5}\Bigg(\frac{1}{2}\Bigg)^{6}$
$=6\cdot\frac{1}{64}$
$=\frac{3}{32}$
- P(at least 5 successes) = P(X ≥ 5)
$=\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$
$=\ ^6\text{C}_\text{5}\Bigg(\frac{1}{2}\Bigg)^{6}\ +\ ^6\text{C}_\text{6}\Bigg(\frac{1}{2}\Bigg)^{6}$
$=6\cdot\frac{1}{64}+1\cdot\frac{1}{64}$
$=\frac{7}{64}$
- P(at most 5 successes) = P(X ≤ 5)
$=1-\text{P}(\text{X}>5)$
$=1-\text{P}(\text{X}=6)$
$=1-\ ^6\text{C}_\text{6}\Bigg(\frac{1}{2}\Bigg)^{6}$
$=1-\frac{1}{64}$
$=\frac{63}{64}$ View full question & answer→