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MATHS 5 Marks Questions

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 275 questions.

Questions · Page 3 of 6

5 Marks Questions

Question 1015 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 1025 Marks
By examining the chest $X$ ray, the probability that $TB$ is detected when a person is actually suffering is $0.99. $ The probability of an healthy person diagnosed to have $TB$ is $0.001.$ In a certain city, $1$ in $1000$ people suffers from $TB.$ A person is selected at random and is diagnosed to have $TB.$ What is the probability that he actually has $TB$?
Answer
Let $E_1 =$ Event that person has $TB$
$E_2 =$ Event that the person does not have $TB$
$E =$ Event that the person is diagnosed to have $TB$
$\therefore\text{P}(\text{E}_1)=\frac{1}{1000}=0.001,$
$\text{P}(\text{E}_2)=\frac{999}{1000}=0.999$
And $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=0.099$ and $\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=0.001$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.001}$
$=\frac{0.000990}{0.000990+0.000999}$
$=\frac{990}{1989}=\frac{110}{221}$
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Question 1035 Marks
Assume that the chances of a patient having a heart attack is $40\%$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Answer
Let $E_1$ and $E_2$ be the events
$E_1 :$ Treatment of yoga and meditation
$E_2 :$ Treatment of prescription of certain drugs
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
Let $A$ denotes that a person has heart risk attack
$\therefore\ \text{P}(\text{A})=40\%=0.40$
Yoga and mediation reduces the heart risk by $30\%$
i. e. inspite of getting yoga and meditation treatment heart risk is $70\%$ of the $0.40$
$\therefore\ \text{P}(\text{A|}\text{E}_1)=0.40\times0.70=0.28$
Drug prescription reduces the heart risk by $25\%$
Even after adopting the drug prescription heart risk is $75\%$ of the $0.40$
$\therefore\ \text{P}(\text{A|E}_2)=0.40\times0.75=0.30$
$\therefore\ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A|E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+0.30\frac{1}{2}}=\frac{0.28}{0.28+0.30}=\frac{28}{58}=\frac{14}{29}$
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Question 1045 Marks
Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.
Answer
Two cards are drawn simultaneously from a pack of 52 cards. Let X denotes the number of kings drawn. So, X = 0, 1, 2. P(X = 0) $=\frac{\text{}^{48}\text{C}_2}{\text{}^{52}\text{C}_2}$ $=\frac{1128}{1326}$ $=\frac{188}{221}$ P(X = 1) $=\frac{\text{}^{4}\text{C}_1\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_2}$ $=\frac{192}{1326}$ $=\frac{32}{221}$ P(X = 2) $=\frac{\text{}^{4}\text{C}_2}{\text{}^{52}\text{C}_2}$ $=\frac{6}{1326}$ $=\frac{1}{221}$ So,
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{x}_\text{i}\text{p}_\text{i}$ $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$ $\frac{188}{221}$ $0$ $0$
$1$ $\frac{32}{221}$ $\frac{32}{221}$ $\frac{32}{221}$
$2$ $\frac{1}{221}$ $\frac{2}{221}$ $\frac{4}{221}$
    $\sum\text{xp}=\frac{34}{221}$ $\sum\text{x}^2\text{p}$
Mean $=\sum\text{xp}$ Mean $=\frac{34}{221}$ Variance $=\sum\text{x}^2\text{p}-(\text{Mean})^2$ $=\frac{36}{221}-\Big(\frac{34}{221}\Big)^2$ $=\frac{7956-1156}{48841}$ $=\frac{6800}{48841}$ Variance $=\frac{400}{2873}.$
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Question 1055 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 1065 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 1075 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 1085 Marks
A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Answer
Let P denote the probability of getting a ticket bearing number divisibal by 10, So
$\text{p}=\frac{10}{100}$ [Since there are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 which are divisible by 10] 
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable representing the number of tickets bearing a number divisible by 10 out of 5 tickets. probability of getting r tickets bearing a number divisible by 10 out ot n tickets is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{\text{5}}\text{c}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}\dots(1)$
Probability of getting all the tickets bearing a number divisible by 10
$=\text{ }^5\text{c}_5\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^{5-5}$ [Using (1)]
$=1.\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^0$
$=\big(\frac{1}{10}\big)^5$
Required probability $=\big(\frac{1}{10}\big)^5$
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Question 1095 Marks
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
Answer
When a die is thrown, probability of getting a six (p) $=\frac{1}{6}$ Therefore, $=\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
  1. If he gets a six in first throw, then,
Probability of getting a six $=\frac{1}{6}$
  1. If he does not get a six, in first throw, but he gets a six in the second throw, then
Its probability $=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$
Probability that he does not get a six in first two throws and he gets a six in thied throw
$=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{216}$
Probability that he does not get a six in any of the three throws $=\bigg(\frac{5}{6}\bigg)^3=\frac{125}{216}$
In first throw he gets a six, will receive Rs.1
If he gets a six in second throw, he will receive Rs.(1 - 1) = 0
If he gets a six in third throw, he will receive Rs.(- 1 - 1 + 1)=Rs. -1 =he will loss Rs. 1
Exapected value $=\frac{1}{6}\times1+\bigg(\frac{5}{6}\times\frac{1}{6}\bigg)\times0+\bigg(\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\bigg)\times(-1)$
$=\frac{1}{6}-\frac{25}{216}=\frac{11}{216}\ (\text{loss})$
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Question 1105 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 1115 Marks
The random variable $X$ can take only the values $0, 1, 2.$ Given that $P(X = 0) = P (X = 1) = p$ and that $E(X^2) = E[X], $ find the value of $p.$
Answer
Since, $X = 0, 1, 2$ and $P (X)$ at $X = 0$ and $1$ is $p,$ let at $X = 2, P (X)$ is $x.$
$\Rightarrow p + p + x = 1$
$\Rightarrow x = 1 – 2p$
We get, the following distribution
$X$ $0$ $1$ $2$
$P(X)$ $p$ $q$ $1 - 2p$
$\therefore\text{E}[\text{X}]=\sum\text{XP}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+2(1-2\text{p})$
$=\text{p}+2-4\text{p}$
$=2-3\text{p}$
And $\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+4\cdot(1-2\text{p})$
$=\text{p}+4-8\text{p}$
$=2-7\text{p}$
Also, given that $\text{E}(\text{X}^2)=\text{E}[\text{X}]$
$\Rightarrow4-7\text{p}=2-3\text{p}$
$\Rightarrow4\text{p}=2$
$\Rightarrow\text{p}=\frac{1}{2}$
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Question 1125 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 1135 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
Answer
Let p be the probability of getting a doublet in a throw of a pair of dice, so
$\text{p}=\frac{6}{36}$ [Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
$=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$ [Since p + q = 1]
$=\frac{5}{6}$
Let X denote the number of grtting doublets i.e. success out of 4 times. So, probility distribution is given by
$\text{X}$ $\text{P(X)}$
$0$ $\text{ }^4\text{C}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{4-0}=\big(\frac{5}{6}\big)^4$
$1$ $\text{ }^4\text{C}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{4-1}=4\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^3=\frac{2}{3}\big(\frac{5}{6}\big)^3$
$2$ $\text{ }^4\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{4-2}=\frac{4.3}{2}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^2=\frac{25}{216}$
$3$ $\text{ }^4\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{4-3}=\frac{4.3}{2}\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)=\frac{5}{324}$
$4$ $\text{ }^4\text{C}_4\big(\frac{1}{6}\big)^4\big(\frac{5}{6}\big)^{4-4}=\big(\frac{1}{6}\big)^4=\frac{1}{1296}$
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Question 1145 Marks
Two cards are drawn simultaneosly from a well shuffled deck of 52 cards. Find the probability distribution of the number of the successes, when getting a spade is considered a success.
Answer
Let X denote the numbers of spades in a sample of two 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 spade)
$=\frac{\text{}^{39}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{741}{1326}$
$=\frac{19}{34}$
P(X = 1)
= P(1 spade)
$=\frac{\text{}^{13}\text{C}_1\times\text{}^{39}\text{C}_1}{\text{}^{52}\text{C}_2}$
$=\frac{507}{1326}$
$=\frac{13}{34}$
P(X = 2)
= P(2 spade)
$=\frac{\text{}^{13}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{78}{1326}$
$=\frac{1}{17}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X})$
$\frac{19}{34}$
$\frac{13}{34}$
$\frac{1}{17}$
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Question 1155 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 1165 Marks
Three dice are thrown at the sametime. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
Answer
As three dice are thrown at the same time, so we have sample space $[n(S)] = 6^3 = 216$
Let $E_1$ is the event when the sum of numbers on the dice was six and $E_2$ is the event when three two’s occurs.
$\Rightarrow E_1 = \{(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3) (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1)\}$
$\Rightarrow n(E_1) = 10$ and $E_2 ={2, 2, 2}$
$\Rightarrow n(E_1) = 1$
Also, $(\text{E}_1\cap\text{E}_2)=1$
$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_2)}$
$=\frac{\frac{1}{216}}{\frac{10}{216}}$
$=\frac{1}{10}$
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Question 1175 Marks
A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.
Answer
It is given that "success" denotes the event of getting the number 1, 3 or 5. Then, P(success) $=\frac{1}{2}$ Also, "failure" denotes the event of getting the numbers 2, 4, or 6. Then, P(failure) $=\frac{1}{2}$ Let X denote the event of getting success. Then, X can take the values 0, 1 and 2. Now, P(X = 0) = P(no success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ P(X = 1) = P(1 success) $=\Big(\frac{1}{2}\times\frac{1}{2}\Big)+\Big(\frac{1}{2}\times\frac{1}{2}\Big)=\frac{1}{2}$ P(X = 2) = P(2 success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ Thus, the probability distribution of X is given by
$\text{X}$ $\text{P}(\text{X})$
$0$ $\frac{1}{4}$
$1$ $\frac{1}{2}$
$2$ $\frac{1}{4}$
Computation of mean and variance
$\text{X}_\text{i}$ $\text{P}_\text{i}$ $\text{P}_\text{i}\text{X}_\text{i}$ $\text{P}_\text{i}\text{X}_\text{i}^2$
$0$ $\frac{1}{4}$ $0$ $0$
$1$ $\frac{1}{2}$ $\frac{1}{2}$ $\frac{1}{2}$
$2$ $\frac{1}{4}$ $\frac{1}{2}$ $1$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=1$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{3}{2}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1$ Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-({\text{Mean}})^2=\frac{3}{2}-1=\frac{1}{2}$
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Question 1185 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer
The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.
Probability of getting doublets in a single throw of the pair of dice is
$\text{p}=\frac{6}{36}=\frac{1}{6}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Clearly, X has the binomial distribution with n = 4, $\text{p}=\frac{1}{6},\ \text{and}\ \text{q}=\frac{5}{6}$
$\therefore\ \text{P}(\text{X}=\text{x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where}\ \text{x}=0,\ 1,\ 2,\ 3...\text{n}$
$=\ ^4\text{C}_\text{x}\Bigg(\frac{5}{6}\Bigg)^{4-\text{x}}. \Bigg(\frac{1}{6}\Bigg)^\text{x}$
$=\ ^4\text{C}_\text{x}\cdot\frac{5}{6^4}^{4-\text{x}}$
∴ P(2 successes) = P(X = 2)
$=\ ^4\text{C}_\text{2}\cdot\frac{5}{6^4}^{4-\text{2}}$
$=6.\frac{25}{1296}$
$=\frac{25}{216}$
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Question 1195 Marks
In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.
Answer
Let X represent the number of correctly answered questions out of 20 questions.
The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.
$\therefore\ \text{p}=\frac{1}{2}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}$
X has a binomial distribution with n = 20 and $\text{p}=\frac{1}{2}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where x}=1,\ 2,\ ...\text{n}$
$=\ ^{20}\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{20-\text{x}}.\bigg(\frac{1}{2}\bigg)^\text{x}$
$=\ ^{20}\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{20}$
P(at least 12 questions answered correctly) = P(X ≥ 12)
$=\text{P}(\text{X}=12)+\text{P}(\text{X}=13)+...+\text{P}(\text{X}=20)$
$=\ ^{20}\text{C}_{12}\bigg(\frac{1}{2}\bigg)^{20}+\ ^{20}\text{C}_{13}\bigg(\frac{1}{2}\bigg)^{20}+...+\ ^{20}\text{C}_{20}\bigg(\frac{1}{2}\bigg)^{20}$
$=\bigg(\frac{1}{2}\bigg)^{20}.\bigg[\ ^{20}\text{C}_{12}+\ ^{20}\text{C}_{13}+...+\ ^{20}\text{C}_{20}\bigg]$
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Question 1205 Marks
A fair coin is tossed four times. Let X denote the longest string of heads accuring. Find the probability distribution mean and variance of X.
Answer
Let the event of getting a head = H and getting a tail = T
Let X denote the variable longest consecutive heads accuring in 4 tosses. The possible values are
X = 0 (no heads) {T, T, T, T}
X = 1 (1 head) {H, T, T, T}
X = 2 (2 heads) {H, H, T, T}
X = 3 (3 heads) {H, H, H, T}
X = 4 (4 heads) {H, H, H, H}
n(S) = {(HHHH), (HHHT), (HHTT), (HTHH), (HTHT), (HTTH), (HTTT), (THHH), (THTH), (THHT), (THTT), (THTT), (TTHH), (TTHT), (TTTH), (TTTT)}
$\text{P}(\text{X}=0)=\frac{1}{16}$
$\text{P}(\text{X}=1)=\frac{7}{16}$
$\text{P}(\text{X}=2)=\frac{5}{16}$
$\text{P}(\text{X}=3)=\frac{2}{16}$
$\text{P}(\text{X}=4)=\frac{1}{16}$
Thus, the probability distribution is
$\text{X}$ $0$ $1$ $2$ $3$ $4$
$\text{p}_\text{i}=\text{P}(\text{X})$ $\frac{1}{16}$ $\frac{7}{16}$ $\frac{5}{16}$ $\frac{2}{16}$ $\frac{1}{16}$
$\text{p}_\text{i}\text{x}_\text{i}^2$ $0$ $\frac{7}{16}$ $\frac{20}{16}$ $\frac{18}{16}$ $1$
Mean $=\sum_\text{i=1 to n}\text{X}_\text{i}\times\text{P}(\text{X}_\text{i})$
Mean, $\mu=0\times\frac{1}{16}+1\times\frac{7}{16}+2\times\frac{5}{16}+3\times\frac{2}{16}+4\times\frac{1}{16}$
$=0+\frac{7}{16}+\frac{10}{16}+\frac{6}{16}+\frac{4}{16}$
$=\frac{27}{16}=1.7$
Variance Var(X) $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{61}{16}-(1.7)^2$
$=3.825-2.89$
$=0.935$
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Question 1215 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 1225 Marks
A coin is tossed 5 times. What is the probability of getting at least 3 heads?
Answer
Probability of getting head on one throw of coin $=\frac{1}{2}$
So, $\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$
$\text{q}=\frac{1}{2}$ [Since p + q = 1]
The coin is tossed 5 times. Let x denote the number of getting head as 5 tosses of coins.
So probability of getting r head in n tosses of coin is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n - r}}$
$\text{P}(\text{x = r})=\text{ }^5\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}}\dots(1)$
probability of getting at least 3 heads
$=\text{P}(\text{X}=3)+\text{P}(\text{x}=4)+\text{p}(\text{x}=5) $
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3.\big(\frac{1}{2}\big)^{5-3}+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{5-4}+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^0$ [Using (1)]
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^2+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5.1$
$=\frac{5.4}{2}.\big(\frac{1}{2}\big)^5+5\big(\frac{1}{2}\big)^5+1.\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5[10+5+1]$
$=16.\frac{1}{32}$
$=\frac{1}{2}$
The required probability is $=\frac{1}{2}$
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Question 1235 Marks
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that:
  1. All the five cards are spades?
  2. Only 3 cards are spades?
  3. None is a spade?
Answer
Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials. In a well shuffled deck of 52 cards, there are 13 spade cards. $\Rightarrow\ \text{p}=\frac{13}{52}=\frac{1}{4}$ $\therefore\ \text{q}=1-\frac{1}{4}=\frac{3}{4}$ X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{4}$ $\text{p}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where x}=0,\ 1,\ ...\text{n}$ $=\ ^5\text{C}_\text{x}\Big(\frac{3}{4}\Big)^{5-\text{x}}\Big(\frac{1}{4}\Big)^\text{x}$
  1. P(all five cards are spades) = P(X = 5)
$=\ ^5\text{C}_{5}\Big(\frac{3}{4}\Big)^0.\Big(\frac{1}{4}\Big)^{5}$
$=1\cdot\frac{1}{1024}$
$=\frac{1}{1024}$
  1. P(only 3 cards are spades) = P(X = 3)
$=\ ^5\text{C}_{3}\cdot\Big(\frac{3}{4}\Big)^2.\Big(\frac{1}{4}\Big)^{3}$
$=10\cdot\frac{9}{16}\cdot\frac{1}{64}$
$=\frac{45}{512}$
  1. P(none is a spade) = P(X = 0)
$=\ ^5\text{C}_{0}\cdot\Big(\frac{3}{4}\Big)^5.\Big(\frac{1}{4}\Big)^{0}$
$=1\cdot\frac{243}{1024}$
$=\frac{243}{1024}$
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Question 1245 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 1255 Marks
A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer
Let $E_{1 }=$ the missing card is a diamond,
$E_{2 }=$ the missing card is a spade,
$E_{3 }=$ the missing card is a club,
$E_{4 }=$ the missing card is a heart and
$A =$ drawing of two heart cards from the remaining cards.
$\text{P}(\text{E}_1)=\frac{13}{52}=\frac{1}{4},\ \text{P}(\text{E}_2)=\frac{13}{52}=\frac{1}{4},$
$\text{P}(\text{E}_3)=\frac{13}{52}=\frac{1}{4},\ \text{P}(\text{E}_4)=\frac{13}{52}=\frac{1}{4}$
$\text{P}(\text{A}|\text{E}_1) = P($ drawing $2$ heart cards given that one diamond card is missing $) =\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}$
Similarly, $ \text{P}(\text{A}|\text{E}_2)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)},\ \text{P}(\text{A}|\text{E}_3)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}$ and $\text{P}(\text{A}|\text{E}_4)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}$
By Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)+\text{P}(\text{E}_4)\text{P}(\text{A}|\text{E}_4)}$
$=\frac{\frac{1}{4}\times\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}}{\frac{1}{4}\times\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}}$
$=\frac{66}{66+78+78+78}=\frac{11}{50}$
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Question 1265 Marks
Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Answer
We know that, in a Bindmial distribution,
  1. There are 2 outcomes for each trial.
  2. There is a fixed number of trials.
  3. The probability of success must be the same for all the trials.
When coin is tossed, possible outcomes are Hesd and Tail.
Since coin is tossed three times, we have fixed number of trials.
Also probility of Head and Tail in each trial is $\frac{1}{2}.$
Thus given experiment is said to have binomial distribution.
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Question 1275 Marks
Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.
Answer
Throwing a doublet i.e. $\big\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\big\}$
Total number of outcomes = 36
Let p be the probability of success therefore
$\text{p}=\frac{6}{3}6=\frac{1}{6}$
Let q be the probability of failure therefore $\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Since there is three rows of dise so n = 3
Let X be the random variable for getting doublet, therefore X can take at max 3 values.
$\text{P(X}=0)=\text{ }^3\text{c}_0\text{p}^0\text{q}^{3}=\big(\frac{5}{6}\big)^{3}=\frac{125}{216}$
$\text{P(X}=1)=\text{ }^3\text{c}_1\text{p}^1\text{q}^2=3.\frac{1}{6}.\big(\frac{5}{6}\big)^2=\frac{75}{216}$
$\text{P(X}=2)=\text{ }^3\text{c}_2\text{p}^2\text{q}^1=3.\big(\frac{1}{6}\big)^2.\big(\frac{5}{6}\big)=\frac{15}{216}$
$\text{P(X}=3)=\text{ }^3\text{c}_3\text{p}^3\text{q}^0=\big(\frac{1}{6}\big)^3=\frac{1}{216}$
Mean
$\mu=\sum^3_{\text{i}-1}\text{X}_{\text{i}}\text{P(X}_{\text{i}})=0.\frac{125}{216}+1.\frac{75}{216}+2.\frac{15}{216}+3.\frac{1}{216}$
$=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2}$
Hence the mean is $=\frac{1}{2}$
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Question 1285 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 1295 Marks
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big).$
Answer
Given,
$\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{3}+\frac{1}{5}-\frac{11}{30}$
$=\frac{10+6-11}{30}$
$=\frac{5}{30}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{6}}{\frac{1}{5}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{6}}{\frac{1}{3}}$
$=\frac{1}{6}\times\frac{3}{1}$
$=\frac{1}{2}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6},\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{1}{2}$
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Question 1305 Marks
A couple has two children. Find the probability that both the children are,
  1. Males, if it is known that at least one of the children is male.
  2. Females, if it is known that the elder child is a female.
Answer
Consider the given events.
$A =$ Both the children are female.
$B =$ The elder child is a female.
$C =$ At least one child is a male.
$D =$ Both children are male.
Clearly,
$\ce{S = \{M_1M_2, M_1F_2, F_1M_2, F_1F_2\}}$
$\ce{A = \{F_1F_2\}}$
$\ce{B = \{F_1M_2, F_1F_2\}}$
$\ce{C = \{M_1F_2, F_1M_2, M_1M_2\}}$
$\ce{D = \{M_1M_2\}}$
$[$Here, first child is elder and second is younger$]$
$\text{D}\cap\text{C}=\big\{\text{M}_1\text{M}_2\big\}$ and $\text{A}\cap\text{B}=\big\{\text{F}_1\text{F}_2\big\}$
  1. Required probability $=\text{P}\Big(\frac{\text{D}}{\text{C}}\Big)=\frac{\text{n}(\text{D}\cap\text{C})}{\text{n}(\text{C})}=\frac{1}{3}$
  2. Required Probability $= \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$
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Question 1315 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 1325 Marks
By examining the chest $X-$ray, probability that $T.B.$ is detected when a person is actually suffering is $0.99.$ The probability that the doctor diagnoses incorrectly that a person has $T.B.$ on the basic of $X-$ray is $0.001.$ In a certain city $1$ in $1000$ persons suffers from $T.B.$ A person is selected at random is diagnosed to have $T.B$. what is the chance that he actually has $T.B.$?
Answer
Consider events $E_1, E_2$ and $A$ as
$E_1 =$ The person selected is actually having $T.B.$
$E_2 =$ The person selected not having $T.B.$
$E_3 =$ The person diagnosed to have $T.B.$
Given,
$\text{P}(\text{E}_1)=\frac{1}{1000}$
$\text{P}(\text{E}_2)=\frac{999}{1000}$
$P(A|E_1) = P($Person diagnosed to have $T.B.$ and he is actually having $T.B.)$
$=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)= P($Person diagnossed to have a $T.B.$ and he is not a actually having $T.B.)$
$0.001$
To find, $P($Person diagnosed to have $T.B.$ is actually having $T.B.) =\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{1000}\times0.99}{\frac{1}{1000}\times0.99+\frac{999}{1000}\times0.001}$
$=\frac{990}{990+999}$
$=\frac{990}{1989}$
$=\frac{110}{221}$
Required probability $=\frac{110}{221}$
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Question 1335 Marks
An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.
Answer
Let X denote the number of successes in 6 trials.
It is given that successes are twice the failures.
$\Rightarrow\text{p}=2\text{q}$
$\text{p + q}=1$
$\Rightarrow3\text{q}=1$
$\Rightarrow\text{q}=\frac{1}{3}$
$\therefore\text{p}=1-\frac{1}{3}=\frac{2}{3}$
$\text{ n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{2}{3}\big)^{\text{r}}\big(\frac{1}{3}\big)^{6-\text{r}},\text{r}=0,1,2,\dots6$
$\text{P(atleast 4 successes})=\text{P(X}\geq4)$
$=\text{P(X}=4)+\text{P(X}=5)+\text{P(X}=6)$
$\text{ }^6\text{C}_4\big(\frac{2}{3}\big)^4\big(\frac{1}{3}\big)^{6-4}+\text{ }^6\text{C}_5\big(\frac{2}{3}\big)^5\big(\frac{1}{3})^{6-5}+\text{ }^6\text{C}_6\big(\frac{2}{3}\big)^6\big(\frac{1}{3}\big)^{6-6}$
$=\frac{15(2^4)+6(32)+64}{3^6}$
$=\frac{240+192+64}{729}$
$=\frac{496}{729}$
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Question 1345 Marks
A bag contains 20 tickets, numbered from 1 to 20. Two tickets are drawn without replacement. What is the probability that the first ticket has an even number and the second an odd number.
Answer
Total number of tickets are 20 numbered from 1, 2, 3, ..... 20.
Number of tickets with even numbers,
= 10 [Since, even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
Number of tickets with odd numbers,
= 10 [Since, odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17,19]
Two cards are drawn without replacement.
A = tickets with even numbers
B = tickets with odd numbers
P (first ticket has even number and second has odd number)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{10}{20}\times\frac{10}{19}$
$=\frac{5}{19}$
Required probability $=\frac{5}{19}$
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Question 1355 Marks
A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.
Answer
A = 4 appears on third toss, if a die is thrown three times
= {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4)
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4)
(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4)
(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)
(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4)
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = 6 and 5 appears respectively on first two tosses, if die is tosses three times
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
$\text{A}\cap\text{B}=\{(6,5,4)\}$
Required probability $=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required Probability $=\frac{1}{6}$
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Question 1365 Marks
Five defective mangoes are acciedently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
Answer
Let X denote number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4. Now, P(X = 0) P(no defective mango) $=\frac{\text{}^{15}\text{C}_4}{\text{}^{20}\text{C}_4}$ $=\frac{1365}{4845}$ $=\frac{91}{323}$ P(X = 1) = P(1 defective mango) $=\frac{\text{}^5\text{C}_1\times\text{}^{15}\text{C}_3}{\text{}^{20}\text{C}_4}$ $=\frac{2275}{4845}$ $=\frac{455}{969}$ P(X = 2) = P(2 defective mangoes) $=\frac{\text{}^5\text{C}_2\times\text{}^{15}\text{C}_2}{\text{}^{20}\text{C}_4}$ $=\frac{1050}{4845}$ $=\frac{70}{323}$ P(X = 2)= P(3 defective mangoes)
$=\frac{\text{}^5\text{C}_3\times\text{}^{15}\text{C}_1}{\text{}^{20}\text{C}_4}$ $=\frac{150}{4845}$ $=\frac{10}{323}$ P(X = 3)= P(4 defective mangoes)
$=\frac{\text{}^5\text{C}_4}{\text{}^{20}\text{C}_4}$ $=\frac{5}{4845}$ $=\frac{1}{969}$ The required probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{91}{323}$
$\frac{455}{969}$
$\frac{70}{323}$
$\frac{10}{323}$
$\frac{1}{969}$
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Question 1375 Marks
An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.
Answer
Urn has 4 red and 3 blue balls. 3 balls are drawn with replacement. Let X denote the numbers of blue balls drawn out of 3 drawn.
So, X has values 0, 1, 2, 3
$\text{P}(\text{X}=0)=\text{P}\big(\overline{\text{B}}_1\big)\times\text{P}\big(\overline{\text{B}}_2\big)\times\text{P}\big(\overline{\text{B}}_3\big)$
$=\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=1)=\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{4}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{7}\times\frac{3}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=2)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{3}{7}\times\frac{4}{7}\times\frac{3}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{108}{343}$
$\text{P}(\text{X}=3)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{27}{343}$
So,
Required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{64}{343}$
$\frac{144}{343}$
$\frac{108}{343}$
$\frac{27}{343}$
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Question 1385 Marks
If the mean and variance of a random variable X having  a binomial distribution are 4 and 2 respectively, find P (X = 1).
Answer
Let n and p be the parmeters of binomial distribution,
Given,
$\text{Mean = np}=4$
$\text{Variance = npq}=2$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$.
$\text{p}=1-\frac{1}{2}$
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X= r})\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}$
$\text{P(X=}1)$
$=\text{ }^8\text{c}_{1}\big(\frac{1}{2}\big)^{1}\big(\frac{1}{2}\big)^{8-1}$
$=8\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^5$
$=\frac{1}{32}$
$\text{P(X}=1)=\frac{1}{32}$
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Question 1395 Marks
Find the variance of the distribution:
$\text{x}$ $0$ $1$ $2$ $3$ $4$ $5$
$\text{P}(\text{x})$ $\frac{1}{6}$ $\frac{5}{18}$ $\frac{2}{9}$ $\frac{1}{6}$ $\frac{1}{9}$ $\frac{1}{18}$
Answer
We have,
$\text{X}$ $0$ $1$ $2$ $3$ $4$ $5$
$\text{P}(\text{X})$ $\frac{1}{6}$ $\frac{5}{18}$ $\frac{2}{9}$ $\frac{1}{6}$ $\frac{1}{9}$ $\frac{1}{18}$
$\text{X}\text{P}(\text{X})$ $0$ $\frac{5}{18}$ $\frac{4}{9}$ $\frac{1}{2}$ $\frac{4}{9}$ $\frac{5}{18}$
$\text{X}^2\text{P}(\text{X})$ $0$ $\frac{5}{18}$ $​​\frac{8}{9}$ $\frac{3}{2}$ $\frac{16}{9}$ $\frac{25}{18}$
$\therefore\text{Variance}=\text{E}(\text{X}^2)-\big[\text{E}(\text{X})\big]^2$
$=\sum\text{X}^2\text{P}(\text{X})-\big[\sum\text{X}\text{P}(\text{X})\big]^2$
$=\Big[0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18}\Big]-\Big[0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18}\Big]^2$
$=\frac{105}{18}-\frac{35^2}{18^2}$
$=\frac{18\times105-35\times35}{18^2}$
$=\frac{19\times35}{324}$
$=\frac{665}{324}$
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Question 1405 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$ then find $\text{P}(\text{A}|\text{B}), \text{ P}(\text{B}|\text{A}), \text{ P}(\overline{\text{A}}|\text{B})$ and $\text{P}(\overline{\text{A}}|\overline{\text{B}}).$
Answer
We have,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$
Also, $\text{P}(\overline{\text{B}})=1-\text{P(B)}=1-\frac{1}{3}=\frac{2}{3}$
As, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$=\frac{6+4-3}{12}$
$\Rightarrow\ \text{P}(\text{A}\cup\text{B})=\frac{7}{12}$
Also, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{3}-\frac{1}{4}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{4-3}{12}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{12}$
And, $\Rightarrow\ \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$=1-\text{P}({\text{A}\cup\text{B}})$
$=1-\frac{7}{12}$
$=\frac{5}{12}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{4},$
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{2}\Big)}=\frac{2}{4}=\frac{1}{2},$
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{1}{12}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{12}=\frac{1}{4}$ and
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}=\frac{\Big(\frac{5}{12}\Big)}{\Big(\frac{2}{3}\Big)}=\frac{15}{24}=\frac{5}{8}$
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Question 1415 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 1425 Marks
Coloured balls are distributed in four boxes as shown in the following table:
Box
Colour
Black
White
Red
Blue
$I$
$II$
$III$
$IV$
$3$
$2$
$1$
$4$
$4$
$2$
$2$
$3$
$5$
$2$
$3$
$1$
$6$
$2$
$1$
$5$
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box $III.$
Answer
Let $A, E_1, E_2, E_3$ and $E_4$ denote the events that the ball is black, box $I$ selected, box $II$ selected, box $III$ is selected and box $IV$ is selected respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{18}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_4}\Big)=\frac{4}{13}$
Using Bayes' theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{4}\times\frac{1}{7}}{\frac{1}{4}\times\frac{3}{18}+\frac{1}{4}\times\frac{2}{8}+\frac{1}{4}\times\frac{1}{7}+\frac{1}{4}\times\frac{4}{13}}$
$=\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{1}{13}}$
$=\frac{156}{947}$
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Question 1435 Marks
One bag contains $4$ yellow and $5$ red balls. Another bag contains $6$ yellow and $3$ red balls. $A$ ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.
Answer
Bag $I$ contains $4$ yellow and $5$ red balls
Bag $II$ contains $6$ yellow and $3$ red balls
Transfer can be done in two ways:
$I - A$ red ball is transferred from bag $I$ to bag $II$ and then one yellow ball is drawn from bag $II.$
$II - A$ red ball is transferred from bag $I$ to bag $II$ and then one yelllow ball is drawn from bag $II.$
Let $E_1, E_2$ and A be events as:
$E_1 =$ One yellow ball drawn from bag $I$
$E_2 =$ One red ball drawn from bag $I$
$A =$ one yellow ball draw from bag $II.$
$\text{P}(\text{E}_1)=\frac{4}{9}$
$\text{P}(\text{E}_2)=\frac{4}{9}$
$\text{P}(\text{A}|\text{E}_1)=\frac{7}{10}$
$[$Since $E_1$ has increased one yellow ball in bag $II]$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{6}{10}$
$[$Since $E_2$ has increased one red ball in bag $II]$
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{4}{9}\times\frac{7}{10}\times\frac{5}{9}\times\frac{6}{10}$
$=\frac{28+30}{90}$
$=\frac{58}{90}$
$=\frac{29}{45}$
Required probability $=\frac{29}{45}$
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Question 1445 Marks
The probability that A hits a target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$, What is the probability that the target will be hit, if each one of A and B shoots at the target?
Answer
Given,
Probability that A hits a target $=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}=\frac{1}{3}$
Probability that B hits the targer $=\frac{2}{5}$
$\Rightarrow\ \text{P(B)}=\frac{2}{5}$
P (Target will be hit)
= 1 - P (target will not be hit)
= 1 - P (Niether A non B hits the target)
$=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P}(\overline{\text{B}})$
$=1-[1-\text{P(A)}][1-\text{P}(\overline{\text{B}})]$
$=1-\Big[1-\frac{1}{3}\Big]\Big[1-\frac{2}{5}\Big]$
$=1-\frac{2}{3},\frac{3}{5}$
$=1-\frac{2}{5}$
$=\frac{2}{5}$
Required probability $=\frac{2}{5}$
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Question 1455 Marks
An urn contains 5 red and 2 blcak balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.
Answer
X can assume values 0 to 2. Yes X is a random variable. P(X = 0) = (Probability of getting no black ball) $=\frac{\text{}^{2}\text{C}_0\times\text{}^{5}\text{C}_2}{\text{}^{7}\text{C}_2}=\frac{1\times\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 1) = (Probability of getting one black ball) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{5}\text{C}_1}{\text{}^{7}\text{C}_2}=\frac{1\times5}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 2) = (Probability of getting two black balls) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{5}\text{C}_0}{\text{}^{7}\text{C}_2}=\frac{1\times1}{\frac{7\times6}{2\times1}}=\frac{2}{42}$ Thus, Probability distribution of random variable X is
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X}) $
$\frac{20}{42}$
$\frac{20}{42}$ 
$\frac{2}{42}$
 
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{X}_\text{i}^2$
$0$ $\frac{20}{42}$ $0$ $0$
$1$ $\frac{20}{42}$ $\frac{20}{42}$ $\frac{20}{42}$
$2$ $\frac{2}{42}$ $\frac{4}{42}$ $\frac{8}{42}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{2}{3}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{2}{3}-\Big(\frac{4}{7}\Big)^2=\frac{50}{147}$
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Question 1465 Marks
A bag contains $3$ white and $2$ black balls and another bag contains $2$ white and $4$ black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.
Answer
Bag $I$ contains $3$ white and $2$ black balls
Bag $II$ contains $2$ white and $4$ black balls
One bag is chosen at random, then one ball is drawn and its is while.
Let $E_1, E_2$ and $A$ be events as:
$E_1 =$ Selecting bag $I$
$E_2 =$ Selecting bag $II$
$A =$ Drawing one white ball
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
$[$Since there are only $2$ bags$]$
$P(A|E_1) = P($Drawing a white ball from bag $I)$
$=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P} [$Drawing a white ball from bag $II]$
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{2}{6}$
$=\frac{3}{10}+\frac{2}{12}$
$=\frac{18+10}{60}$
$=\frac{28}{60}$
$=\frac{7}{15}$
Required probability $=\frac{7}{15}$
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Question 1475 Marks
A and B are two independent events. The probability that A and B occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Find the probability of occurrence of two events.
Answer
Given
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{3}$
We know that,
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$\frac{1}{3}=(1-\text{P(A)})(1-\text{P(A)})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P(A) }\text{P(B)}$
$\frac{1}{3}=1-\text{P(B)}-\text{P(A)}+\text{P}(\text{A}\cap\text{B})$
$\frac{1}{3}=1-\text{P(B)}-\text{P(B)}+\frac{1}{6}$
$\text{P(A)}+\text{P(A)}=\frac{1}{1}+\frac{1}{6}-\frac{1}{3}$
$=\frac{6+1-2}{6}$
$\text{P(A)}+\text{P(B)}=\frac{5}{6}$
$\text{P(A)}=\frac{5}{6}-\text{P(B)}\ .....\text{(i)}$
Given, $\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P(A) } \text{P(B)}=\frac{1}{6}$
$\Big[\frac{5}{6}-\text{P(B)}\Big]\text{P(B)}=\frac{1}{6}$
[Using equation (i)]
$\Rightarrow\ \frac{5}{6}\text{P(B)}-\big\{\text{P(B)}\big\}^2=\frac{1}{6}$
$\Rightarrow\ \{\text{P(B)}\}^2-\frac{5}{6}\text{P(B)}+\frac{1}{6}=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-5\text{P(B)}+1=0$
$\Rightarrow\ 6\{\text{P(B)}\}^2-3\text{P(B)}-2\text{P(B)}+1=0$
$\Rightarrow\ 3\text{P(B)}[2\text{P(B)}-1]-1[2\text{P(B)}-1]=0$
$\Rightarrow\ [2\text{P(B)}-1][3\text{P(B)-1}]=0$
$\Rightarrow\ 2\text{P(B)}-1 = 0\text{ or }3\text{P(B)}-1=0$
$\Rightarrow\ \text{P(B)}=\frac{1}{2} \text{ or P(B)}=\frac{1}{3}$
⇒ Using equation (i),
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$
$\text{P(B)}=\frac{1}{2}\Rightarrow\ \text{P(A)}=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}$
Hence, $\text{P(B)}=\frac{1}{2},\text{P(A)}=\frac{1}{3} \text{ or }\text{P(B)}=\frac{1}{3},\text{P(A)}=\frac{1}{2}$
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Question 1485 Marks
A bag contains 3 red and 2 black balls. One ball is drawn from it at random. Its colour is noted and then it is put back in the bag. A second draw is made and the same procedure is repeated. Find the probability of drawing,
  1. Two red balls,
  2. Two black balls,
  3. First red and second black ball.
Answer
Given bag contains 3 red and 2 black balls. A = Getting one red ball $\Rightarrow\ \text{P(A)}=\frac{3}{5}$ B = Getting one black ball $\Rightarrow\ \text{P(B)}=\frac{2}{5}$
  1. P(Getting two red balls)
= P(A) P(A)
$=\frac{3}{5}\times\frac{3}{5}$
$=\frac{9}{25}$
P(Getting two red balls) $=\frac{9}{25}$
  1. P(Getting two black balls)
= P(B) P(B)
$=\frac{2}{5}\times\frac{2}{5}$
$=\frac{4}{25}$
P(Getting two black balls) $=\frac{4}{25}$
  1. P(Getting first red and second black ball)
= P(A) P(B)
$=\frac{3}{5}\times\frac{2}{5}$
$=\frac{6}{25}$
P(Getting first red and second black ball) $=\frac{6}{25}$
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Question 1495 Marks
cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of red cards.
Answer
It is given that the cards are drawn successively with replacement so the events are indepent. Therefore, the drawing of the cards follow binomial distribution. Probability of drawng a red card $=\text{p}=\frac{26}{52}=\frac{1}{2}$ $\therefore\text{q}=1-\text{p}=1-\frac{1}{2}=-\frac{1}{2}$ Also, $\text{n}=3$ Let X be the random variable denoting the number of red cards drawn from a well shuffied pack of 52 cards. $\therefore\text{P(X = r})=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{3-\text{r}}\big(\frac{1}{2}\big)^{\text{r}}=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^3,\text{r}=0,1,2,3$ Probability of drawing no red ball $=\text{P(X}=0)=\text{ }^3\text{C}_0\big(\frac{1}{2}\big)^3=\frac{1}{8}$ Probability of drawing one red ball $=\text{P(X}=1)=\text{ }^3\text{C}_1\big(\frac{1}{2}\big)^3=\frac{3}{8}$ Probability of drawing tow red balls $=\text{P(X}=2)=\text{ }^3\text{C}_2\big(\frac{1}{2}\big)^3=\frac{3}{8}$ Probability of drawing three red balls $=\text{P(X}=3)=\text{ }^3\text{C}_3\big(\frac{1}{2}\big)^3=\frac{1}{8}$ Thus, the probability distribution of X is as follows:
$\text{x}_{\text{i}}$ $\text{p}_{\text{i}}$ $\text{p}_{\text{i}}\text{x}_{\text{i}}$ $\text{p}_{\text{i}}\text{x}_{\text{i}}^2$
$0$ $\frac{1}{8}$ $0$ $0$
$1$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{3}{8}$
$2$ $\frac{3}{8}$ $\frac{6}{8}$ $\frac{12}{8}$
$3$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{9}{8}$
    $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}$ $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2=3$
Mean of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}=\frac{3}{2}$ variance of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2-(\text{Mean})^2=3-\frac{9}{4}=\frac{3}{4}$
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Question 1505 Marks
Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.
Answer
A pair of dice are thrown. It has 36 elem ents in its samplw space.
A = Occurence of number 4 on firs die
A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
B = Occurence of 5 on second die
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
$\text{A}\cap\text{B}=\Big\{(4,5)\Big\}$
$\text{P(A)}=\frac{6}{36}=\frac{1}{6}$
$\text{P(B)}=\frac{6}{36}=\frac{1}{6}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}\frac{1}{6}\times\frac{1}{6}$
$=\frac{1}{36}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are indepepndent events.
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