Chapter 4

A coin is tossed twice. Find the probability distribution of the number of heads.

A die is thrown twice. In the throw getting odd numbers is taken as a success. Find the probability distribution of the success. [MP-2001]

For a biased die, the probabilities for different faces to turn up are \(\begin{array}{lcccccc}\text { Face } & : 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Probability: } & 0.2 & 0.22 & 0.11 & 0.25 & 0.05 & 0.17\end{array}\) The die is tossed and you are told that either face 4 or face 5 has turned up. The probability that it is face 4 is (a) \(1 / 6\) (b) \(1 / 4\) (c) \(5 / 6\) (d) None

Two cards are drawn successively with replacement from well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. [CBSE-1995, 2001]

There are three urns \(A, B\) and \(C\). Urn \(A\) contains 4 white balls and 5 blue balls. \(\operatorname{Urn} B\) contains 4 white balls and 3 blue balls. \(\operatorname{Um} C\) contains 2 white balls and 4 blue balls. One ball is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white balls and one is a blue ball. [CBSE-96]

Four boys and three girls stand in a queue for an interview, probability that they will stand in alternate position is (a) \(1 / 34\) (b) \(1 / 35\) (c) \(1 / 17\) (d) \(1 / 68\)

Word UNIVERSITY is arranged randomly. Then the probability that both \(I\) does not come together is: [UPSEAT-2001] |(a) \(3 / 5\) (b) \(2 / 5\) (c) \(4 / 5\) (d) \(1 / 5\) Solution (c) Total number of ways \(=\frac{10 !}{2 !}\) Favourable number of ways for \(I\) come together is \(9 !\) Thus, probability that \(I\) come together $$ =\frac{9 ! \times 2 !}{10 !}=\frac{2}{10}=\frac{1}{5} $$

Tickets are numbered from 1 to \(10 .\) Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and on the other a multiple of 4 [CBSE-94]

A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from another bag. Find the probability that the ball drawn is white. [CBSE-94]

If four vertices of a regular octagon are chosen at random, then the probability that the quadrilateral formed by them is a rectangle is (a) \(1 / 8\) (b) \(2 / 21\) (c) \(1 / 32\) (d) \(1 / 35\)

Let \(S\) be a set containing \(n\) elements. If we select two subsets \(A\) and \(B\) of \(S\) at random then the probability that \(A \cup B=S\) and \(A \cap\) \(B=\phi\) is: [Orissa JEE 2005] (a) \(2^{n}\) (b) \(n^{2}\) (c) \(1 / n\) (d) \(1 / 2^{n}\) Solution (d) Ways of selecting two subsets of \(A=\left(2^{n}\right)^{2}\) Ways of selecting \(A \cup B\) and \(A \cap B\) are \(2^{n}$$\therefore\) Required probability \(=\) $$ \frac{\text { Favourable cases }}{\text { Total cases }}=\frac{2^{n}}{\left(2^{n}\right)^{2}}=\frac{1}{2^{n}} $$

A bag contains 5 red and 7 black balls. Second bag contains 4 blue and 3 green balls. One ball is drawn from each bag. Find the probability for: \([\) MP-2005 (B)] (i) 1 red and 1 blue ball (ii) 1 green and 1 black ball

In a class \(30 \%\) students fail in physics, \(25 \%\) fail in maths and \(10 \%\) fail in both. A student is chosen at random. Find the probability that (a) He fails in maths if fail in physics. (b) He fails in physics if he has been failed in maths. (c) He is fail in maths or physics.

There are three bags which are known to contain 2 white and 3 black; 4 white and 1 black and 3 white and 7 black balls, respectively. A ball is drawn at random from one of the bags and found to be a black ball. The probability that it was drawn from the bag containing the most black balls is (a) \(7 / 15\) (b) \(5 / 19\) (c) \(3 / 4\) (d) None of these

A die is tossed thrice. A success is getting 1 or 6 on a toss. The mean and the variance of number of successes are: (a) \(\mu=1, \sigma^{2}=2 / 3\) (b) \(\mu=2 / 3, \sigma^{2}=1\) (c) \(\mu=2, \sigma^{2}=2 / 3\) (d) None of theseSolution (a) For binomial distribution, mean \(=n p\) and $$ \begin{aligned} &\text { variance }=n p q \\ &n=3, p=\frac{2}{6}=\frac{1}{3}, q=1-p=1-\frac{1}{3}=\frac{2}{3} \\ &\text { So, } \operatorname{mean}(\mu)=3 \times \frac{1}{3}=1 \\ &\begin{aligned} \text { Variance }\left(\sigma^{2}\right) &=3 \times \frac{1}{3} \times \frac{2}{3} \\ &=\frac{2}{3} \end{aligned} \end{aligned} $$

In a bulb factory, machines \(A, B\) and \(C\) manufacture \(60 \%, 30 \%\) and \(10 \%\) bulbs, respectively. \(1 \%, 2 \%\) and \(3 \%\) of the bulbs produced, respectively, by \(A, B\) and \(\mathrm{C}\) are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by the machine \(A\). [CBSE-2008]

In a certain town, \(40 \%\) of the people have brown hair, \(25 \%\) have brown eyes and \(15 \%\) have both brown hair and brown eyes. If a person selected at random from the town has brown hair, the probability that he also has brown eyes is (a) \(1 / 5\) (b) \(3 / 8\) (c) \(1 / 3\) (d) \(2 / 3\)

To form a committee of 4 persons from 5 women and 7 men, find the probability when the committee contains: (i) 3 women and 1 men (ii) 2 women and 2 men (iii) 4 women

Solution (a) For binomial distribution, mean \(=n p\) and $$ \begin{aligned} &\text { variance }=n p q \\ &n=3, p=\frac{2}{6}=\frac{1}{3}, q=1-p=1-\frac{1}{3}=\frac{2}{3} \\ &\text { So, } \operatorname{mean}(\mu)=3 \times \frac{1}{3}=1 \\ &\begin{aligned} \text { Variance }\left(\sigma^{2}\right) &=3 \times \frac{1}{3} \times \frac{2}{3} \\ &=\frac{2}{3} \end{aligned} \end{aligned} $$

\(A, B\) and \(C\) are three horses participating in a race. The chance of \(A\) 's win is double of \(B\) and chance of \(B\) 's win is double of \(C\). Find out the probability for winning of each of them. Also find the probability that which horse wins the race, \(B\) or \(C\). [MP-2008]

In a group of students, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.

A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is (a) \(3 / 4\) (b) \(1 / 2\) (c) \(1 / 3\) (d) None

The value of \(C\) for which \(P(X=k)=C k^{2}\) can serve as the probability function of a random variable \(X\) that takes \(0,1,2,3,4\) is [EAMCET-1994] (a) \(1 / 30\) (b) \(1 / 10\) (c) \(1 / 3\) (d) \(1 / 15\) (a) \(\sum_{k=0}^{4} P(X=k)=1 \Rightarrow \sum_{k=0}^{4} C_{k}^{2}=1\) \(\Rightarrow C\left(1^{2}+2^{2}+3^{2}+4^{2}\right)=1\) \(\Rightarrow C=\frac{1}{20}\)Solution (b) Here mean \(=n p\) and variance \(=n p q\) $$ \begin{gathered} \therefore \frac{P(X=k)}{P(X=k-1)}=\frac{{ }^{n} C_{k}(p)^{k}(q)^{n-k}}{{ }^{n} C_{k-1}(p)^{k-1}(q)^{n-k-1}}=\frac{{ }^{n} C_{k}}{{ }^{n} C_{k-1}} \times \frac{p}{q} \\ \quad \therefore \frac{P(X=k)}{P(X=k-1)}=\frac{n-k+1}{k} \times \frac{p}{q} \end{gathered} $$ 7\. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards then the mean of the number of aces is (a) \(1 / 13\) (b) \(3 / 13\) (c) \(2 / 13\) (d) None of these Solution (c) Let \(X\) denote a random variable which is the number of aces. Clearly, \(X\) takes values 1,2 . $$ \begin{aligned} &\therefore p=\frac{4}{52}=\frac{1}{13}, q=1-\frac{1}{13}=\frac{12}{13} \\ &P(X=1)=2 \times\left(\frac{1}{13}\right)\left(\frac{12}{13}\right)=\frac{24}{169} \\ &P(X=2)=2 \times\left(\frac{1}{13}\right)^{2}\left(\frac{12}{13}\right)^{0}=\frac{1}{169} \\\ &\text { Mean }=\sum P_{i} X_{i}=\frac{24}{169}+\frac{2}{169}=\frac{26}{169}=\frac{2}{13} \end{aligned} $$

Twelve cards, numbered 1 to 12 , are placed in a box, mixed up throughly. Then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3 , then find the probability that it is an even number. [CBSE-2008]

There are 3 red and 5 black balls in bag. \(A\) and 2 red and 3 black balls in bag \(B\). One ball is drawn from bag \(A\) and two from bag \(B\). Find the probability that out of the 3 balls drawn, 1 is red and 2 are black. [CBSE-96 (C)]

A bag contains 3 white, 3 black and 2 red balls. One by one three balls are drawn without replacing them. The probability that the third ball is red is (a) \(1 / 2\) (b) \(1 / 3\) (c) \(2 / 3\) (d) \(1 / 4\)

Tickets are marked from 1 to 12 and mixed up. One ticket is taken out at random. Find the probability of its being a multiple of 2 or 3 . [MP-91, 94, 2000, 2009]

Two cards are drawn at random from a wellshuffled pack of 52 cards. What is the probability that either both are red or both are kings?

A determinant is chosen at random. The set of all determinants of order 2 with elements 0 or 1 only. The probability that value of the determinant chosen is positive is (a) \(3 / 16\) (b) \(3 / 8\) (c) \(1 / 4\) (d) None

A sample of 4 items is drawn at a random without replacement from a lot of 10 items. Containing 3 defective. If \(X\) denotes the number of defective items in the sample then \(P(0

Out of 21 tickets marked with numbers from 1 to 21 , three are drawn at random. The chance that the numbers on them are in A.P., is (a) \(10 / 133\) (b) \(9 / 133\) (c) \(9 / 1330\) (d) None of these

A sample of 4 items is drawn at a random without replacement from a lot of 10 items. Containing 3 defective. If \(X\) denotes the number of defective items in the sample then \(P(0

A bag \(x\) contains 3 white balls and 2 black balls and another bag \(y\) contains 2 white balls and 4 black balls. A bag and a ball out of it are picked at random. The probability that the ball is white is (a) \(3 / 5\) (b) \(7 / 15\) (c) \(1 / 2\) (d) None

A biased die is tossed and the respective probabilities for various faces to turn up are given below: \(\begin{array}{lcccccc}\text { Face } & : 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Probability: } & 0.1 & 0.24 & 0.19 & 0.18 & 0.15 & 0.14\end{array}\) If an even face has turned up, then the probability that it is face 2 or face 4 is (a) \(0.25\) (b) \(0.42\) (c) \(0.75\) (d) \(0.9\)

\(A\) has 3 shares in a lottery containing 3 prizes and 9 blanks. \(B\) has two shares in a lottery containing 2 prizes and 6 blanks; Find theratio of \(A\) 's chance of success to \(B\) 's chance of success. (a) \(927: 715\) (b) \(972: 751\) (c) \(925: 715\) (d) \(715: 972\) olution (c) Since \(A\) has 3 shares in a lottery, his chance of success means that he gets at least 1 prize, that is, he gets either 1 prize or 2 prizes or 3 prizes and his chance of failure means that he gets no prize. It is certain that either he succeeds or fails. If \(p\) denotes his chance of success and \(q\) the chance of his failure, then \(p+q=1\) or \(p=1-q\) We now find \(q \times n=\) total number of ways $$ ={ }^{12} C_{3}=\frac{12 \times 11 \times 10}{1 \times 2 \times 3}=220 $$ Since out of 12 tickets in the lottery, he can draw any 3 tickets by virtue of his having 3 shares in the lottery and \(m=\) favourable number of ways $$ ={ }^{9} C_{3}=\frac{9 \times 8 \times 7}{1 \times 2 \times 3}=84 $$ Since he will fail to draw a prize if all the tickets drawn by him are blanks. $$ \therefore q=\frac{m}{n}=\frac{84}{220}=\frac{21}{55} $$ \(\therefore p=A\) 's chance of success \(=1-\frac{21}{55}=\frac{34}{55}\)Similarly B's chance of success $$ \begin{aligned} p^{\prime} &=1-q^{\prime}=1-\frac{{ }^{6} C_{2}}{{ }^{8} C_{2}}=1-\frac{6 \times 5}{8 \times 7} \\ &=1-\frac{15}{28}=\frac{13}{28} \end{aligned} $$ \(\therefore A\) 's chance of success: B's chance of success $$ =p: p^{\prime}=\frac{34}{35}: \frac{13}{28}=\frac{952}{1540}: \frac{715}{1540}=952: 715 $$

Two squares are chosen at random on a chessboard. The probability that they have a side in common is (a) \(1 / 9\) (b) \(2 / 7\) (c) \(1 / 18\) (d) None of these

\(A\) and \(B\) throw a pair of dice. \(A\) wins if he throws 6 before \(B\) throws 7 and \(B\) wins if hethrows 7 before \(A\) throws 6 . If \(A\) begins, what is his chance of winning? (a) \(30 / 61\) [MNR 1995] (c) \(61 / 30\) (1) olution (a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of \(B\) 's throwing 7 with a pair of dice. Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly there are six ways of getting 7 , namely, \((1,6)\), \((2,5),(3,4),(4,3),(5,2),(6,1)\) $$ \therefore P\left(E_{1}\right)=\frac{5}{36} $$ and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\) $$ P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6} $$ and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\) It is given that \(A\) starts the game and he will win in the following mutually exclusive ways. (i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw. (ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw. (iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and \(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text {th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\) \(+\ldots\)throws 7 before \(A\) throws 6 . If \(A\) begins, what is his chance of winning? (a) \(30 / 61\) [MNR 1995] (c) \(61 / 30\) (1) olution (a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of \(B\) 's throwing 7 with a pair of dice. Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly there are six ways of getting 7 , namely, \((1,6)\), \((2,5),(3,4),(4,3),(5,2),(6,1)\) $$ \therefore P\left(E_{1}\right)=\frac{5}{36} $$ and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\) $$ P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6} $$ and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\) It is given that \(A\) starts the game and he will win in the following mutually exclusive ways. (i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw. (ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw. (iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and \(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text {th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\) \(+\ldots\)

There are 4 envelopes with addresses and 4 concerning letters. The probability that letter does not go into concerning proper envelope is (a) \(19 / 24\) (b) \(21 / 23\) (c) \(23 / 24\) (d) \(1 / 24\)

Out of \((2 n+1)\) consecutively numbered tickets, three are drawn at random. The chance that the numbers on them are in A.P. is (a) \(4 n^{2}-1 / 3 n\) (b) \(4 n^{2}+1 / 3 n\) (c) \(3 n / 4 n^{2}-1\) (d) \(3 n / 4 n^{2}+1\) (b) If the smallest number is 1 , the groups of three numbers in A.P. are as \(1,2,3 ; 1,3\), \(5 ; 1,4,7 ; \ldots ; 1, n+1,2 n+1 ;\) and they are \(n\) in number. If the smallest number selected is 2, the possible groupings are \(2,3,4 ; 2,4,6 ; 2,5,8 ; \ldots ; 2, n+1,2 n\) and their number is \(n-1\). If the lowest number is 3, the groupings are \(3,4,5 ; 3,5,7 ; 3,6,9 ; \ldots ; 3, n+2,2 n+1\) their number being \(n-1\). Similarly, it can be seen that if the lowest numbers selected are \(4,5,6,2 n-2,2 n-1\), the numbers of selections, respectively, are \(n-2, n-2, n-3, n-3, \ldots, 2,2,1,1\). Thus, the favourable ways for 2,3 are the same and similarly they are the same for 4,5 and so on. Hence, number of favourably ways $$ \begin{aligned} M &=2(1+2+3+\ldots n-1)+n \\ &=2 \times \frac{(n-1) n}{2}+n=n^{2}-n+n=n^{2} \end{aligned} $$ Also the total number of ways $$ \begin{aligned} N &={ }^{2 n+1} C_{3}=\frac{(2 n+1) \times 2 n \times(2 n-1)}{1 \times 2 \times 3} \\ &=\frac{n\left(4 n^{2}-1\right)}{3} \end{aligned} $$Hence, the required probability $$ =\frac{M}{N}=\frac{3 n^{2}}{n\left(4 n^{2}-1\right)}=\frac{3 n}{4 n^{2}-1} $$

The probability distribution of a random variable \(X\) is given below: \(\begin{array}{llll}X=x_{i} & 2 & 3 & 4 \\ P:\left(X=x_{i}\right) & 1 / 4 & 1 / 8 & 5 / 8\end{array}\) Then its mean is (a) \(27 / 8\) (b) \(5 / 4\) (c) 1 (d) \(4 / 5\)

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(1 / 3\) and the probability that he copies the answer is \(1 / 6\). The probabilitythat his answer is correct given that he copied it, is \(1 / 8\). The probability that he knew the answer to the question given that he correctly answered it is [IIT-1991] (a) \(24 / 29\) (b) \(25 / 24\) (c) \(29 / 24\) (d) \(24 / 25\) Solution (a) Let \(A_{1}\) be the event that the examinee guesses the answer; \(A_{2}\) the event that he copies the answer and \(A_{3}\) the event that he knows the answer. Also let \(A\) be the event that he answers correctly. Then as given, we have $$ P\left(A_{1}\right)=\frac{1}{3}, P\left(A_{2}\right)=\frac{1}{6}, P\left(A_{3}\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2} $$ (We have assumed here that the events \(A_{1}, A_{2}\) and \(A_{3}\) are mutually exclusive and totally exhaustive.) Now \(P\left(A / A_{1}\right)=\frac{1}{4}, P\left(A / A_{2}\right)=\frac{1}{8}\) (as given) Again it is reasonable to take the probability of answering correctly given that he knows the answer as 1 , that is, \(P\left(A / A_{3}\right)=1\) We have to find \(P\left(A_{3} / \mathrm{A}\right)\) By Bayes' theorem, we have $$ \begin{aligned} &P\left(A_{3} / A\right)=\frac{P\left(A_{3}\right) P\left(A / A_{3}\right)}{P\left(A_{1}\right) P\left(A / A_{1}\right)+P\left(A_{2}\right) P\left(A / A_{2}\right)+} \\ &\quad P\left(A_{3}\right) P\left(A / A_{3}\right) \\ &=\frac{(1 / 2) \times 1}{(1 / 3)(1 / 4)+(1 / 6)(1 / 8)+(1 / 2) \times 1}=\frac{24}{29} \end{aligned} $$

The probability for a randomly chosen month to have its 10 th day as Sunday is (a) \(1 / 84\) (b) \(10 / 12\) (c) \(10 / 84\) (d) \(1 / 7\) (e) \(1 / 12\)

In a combat between \(A, B\) and \(C, A\) tries to hit \(B\) and \(C\), and \(B\) and \(C\) try to hit \(A\). Probability of \(A, B\) and \(C\) hitting the targets are \(2 / 3,1 / 2\) and \(1 / 3\), respectively. If \(A\) is hit, find the probability that \(B\) hits \(A\) and \(C\) does not. (a) \(1 / 1\) (b) \(1 / 2\) (c) \(2 / 2\) (d) \(2 / 1\) (c) We have to find the probability of \(A\) being hit by \(B\) but not by \(C\), i.e.,$$ \begin{aligned} P\left(B C^{\prime} / A\right)=& \frac{P\left(A / B C^{\prime}\right) P\left(B C^{\prime}\right)}{P\left(A / B C^{\prime}\right) P\left(B C^{\prime}\right)+P\left(A B C^{\prime}\right) P\left(B^{\prime} C\right)} \\ &+P(A / B C) P(B C)+P\left(A / B^{\prime} C^{\prime}\right) P\left(R^{\prime} C^{\prime}\right) \end{aligned} $$ Now putting the values from the given data, we have

Three letters are drawn from the alphabet of 26 letters without replacement. The probability that they appear in alphabetical order is (a) \({ }^{23} C_{1}{\underline{\phantom{xx}}}^{26} C_{3}\) (b) \(24 /{ }^{26} \mathrm{C}_{3}\) (c) \(1 / 6\) (d) \(1 / 3\)

Two persons \(A\) and \(B\) throw a die alternately till one of them gets 3 and wins the game. If \(A\) begins, then their respective probabilities of winning will be \([\) Kerala (CEE)-05] (a) \(6 / 11,5 / 11\) (b) \(5 / 11,4 / 11\) (c) \(7 / 11,6 / 11\) (d) \(1 / 2,1 / 2\) (a) A will win either in first or in third or in fifth throw. So $$ \begin{aligned} P(A) &=\frac{1}{6}+\left(\frac{5}{6}\right)^{2} \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \frac{1}{6}+\ldots \\ &=\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}\right]+\ldots \\\ &=\frac{1}{6}\left[\frac{1}{1-\frac{25}{36}}\right]=\frac{6}{11} \end{aligned} $$ \(B\) will win in second or in fourth or in sixth throw. So $$ \begin{aligned} P(B) &=\left(\frac{5}{6}\right) \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \frac{1}{6}+\ldots \\ &=\left(\frac{5}{6}\right) \frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots\right] \\\ &=\frac{5}{36} \times \frac{36}{11}=\frac{5}{11} \end{aligned} $$ Hence, required respective probabilities = \(6 / 11,5 / 11\)

A random variable \(X\) has the following probability distribution \(\begin{array}{rl}x: 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ P(X=x) & : a & 3 a & 5 a & 7 a & 9 a & 11 a & 13 a & 15 a & 17 a\end{array}\) Then value of \(a\) is (a) \(1 / 81\) (b) \(2 / 81\) (c) \(5 / 81\) (d) \(7 / 81\)

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters. TA are visible. The probability that the letters have come from CALCUTTA is (a) \(1 / 3\) (b) \(4 / 11\) (c) \(5 / 12\) (d) None of these (b) Let \(A\) : 'the event that letters came from TATANAGAR' B: 'the event that letters came from CALCUTTA' \(C\) : 'the event that two consecutive letters visible by \(T A^{\prime}\) Then \(P(A)=1 / 2, P(B)=1 / 2, P(C / B)=1 / 7\) Hence, by Bayes' theorem $$ P(B / C)=\frac{P(B) \times P(C / B)}{P(A) P(C / A)+P(B) P(C / B)}=\frac{4}{11} $$

In 324 throws of 4 dice, the expected number of times three sixes occur is (a) 81 (b) 5 (c) 9 (d) 31

A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is (a) \(1 / 5\) (b) \(2 / 5\) (c) \(3 / 5\) (d) \(4 / 5\) (b) Let A: 'event that sum is 5 ' \(B:\) 'event that sum is 7' \(C:\) 'event that sum is neither 5 nor 7 ' Then \(P(A)=\frac{4}{36}=\frac{1}{9}, P(B)=\frac{6}{36}=\frac{1}{6}\) \(P(C)=\frac{26}{36}=\frac{13}{18}\) Now probability that \(A\) occurs before \(B\) $$ \begin{aligned} &=P(A+C A+C C A+\ldots) \\ &=P(A)+P(C A)+P(C C A)+\ldots \\ &=P(A)+P(C) P(A)+P(C) P(C) P(A)+\ldots \\ &=\frac{1}{9}+\left(\frac{13}{18}\right) \frac{1}{9}+\left(\frac{13}{18}\right)^{2} \frac{1}{9}+\ldots \\ &=\frac{1 / 9}{1}=\frac{2}{5} \end{aligned} $$

If \(x\) denotes the number of sixes in four consecutive throws of a dice, then \(P(x=4)\) is (a) \(1 / 1296\) (b) \(4 / 6\) (c) 1 (d) \(1295 / 1296\)