Chapter 27

In a game two players \(A\) and \(B\) take turns in throwing a pair of fair dice starting with player \(A\) and total of scores on the two dice, in each throw is noted \(A\) wins the game if he throws a total of 6 before \(B\) throws a total of 7 and \(B\) wins the game if he throws a total of 7 before \(A\) throws a total of six. The game stops as soon as either of the players wins. The probability of \(A\) winning the game is : [Sep. \(04,2020(\mathrm{II})]\) (a) \(\frac{5}{31}\) (b) \(\frac{31}{61}\) (c) \(\frac{5}{6}\) (d) \(\frac{30}{61}\)

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4 . Then the conditional probability that the score 4 has appeared atleast once is : [Sep. 03, 2020 (I)] (a) \(\frac{1}{4}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{8}\) (d) \(\frac{1}{9}\)

The probability that a randomly chosen 5 -digit number is made from exactly two digits is: \(\quad\) [Sep.03, 2020 (II)] (a) \(\frac{135}{10^{4}}\) (b) \(\frac{121}{10^{4}}\) (c) \(\frac{150}{10^{4}}\) (d) \(\frac{134}{10^{4}}\)

Box I contains 30 cards numbered 1 to 30 and Box Il contains 20 cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is: \([\) Sep. 02, 2020 (I)] (a) \(\frac{2}{3}\) (b) \(\frac{8}{17}\) (c) \(\frac{4}{17}\) (d) \(\frac{2}{5}\)

Let \(E^{C}\) denote the complement of an event \(E\). Let \(E_{1}, E_{2}\) and \(\mathrm{E}_{3}\) be any pairwise independent events with \(\mathrm{P}\left(\mathrm{E}_{1}\right)>0\) and \(\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=0\). Then \(\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}} \cap \mathrm{E}_{3}^{\mathrm{C}} / \mathrm{E}_{1}\right)\) is equal to: \([\) Sep. \(\mathbf{0 2}, \mathbf{2 0 2 0}(\mathrm{II})]\) (a) \(\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right)\) (b) \(\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)\) (c) \(\mathrm{P}\left(\mathrm{E}_{3}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}}\right)\) (d) \(\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}\right)\)

In a box, there are 20 cards, out of which 10 are labelled as A and the remaining 10 are labelled as B. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is : [Jan. 9, 2020 (I)] (a) \(\frac{9}{16}\) (b) \(\frac{11}{16}\) (c) \(\frac{13}{16}\) (d) \(\frac{15}{16}\)

Let \(A\) and \(B\) be two independent events such that \(P(A)=\frac{1}{3}\) and \(P(B)=\frac{1}{6} .\) Then, which of the following is \(\begin{aligned}&\text { TRUE? } & \text { [Jan. 8, 2020 (I)] }\end{aligned}\) (a) \(P(A / B)=\frac{2}{3}\) (b) \(P\left(A / B^{\prime}\right)=\frac{1}{3}\) (c) \(P\left(A^{\prime} / B^{\prime}\right)=\frac{1}{3}\) (d) \(P(A /(A \cup B))=\frac{1}{4}\)

An unbiased coin is tossed 5 times. Suppose that a variable \(X\) is assigned the value \(k\) when \(k\) consecutive heads are obtained for \(k=3,4,5\), otherwise \(X\) takes the value \(-1\). Then the expected value of \(X\), is: [Jan. 7, 2020 (I)\\} (a) \(\frac{3}{16}\) (b) \(\frac{1}{8}\) (c) \(-\frac{3}{16}\) (d) \(-\frac{1}{8}\)

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is \(\frac{1}{4}\) If the probability that at most two machines will be out of service on the same day is \(\left(\frac{3}{4}\right)^{3} k\), then \(k\) is equal to: [Jan. 7, 2020 (II)] (a) \(\frac{17}{8}\) (b) \(\frac{17}{4}\) (c) \(\frac{17}{2}\) (d) 4

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \(\frac{4}{5}\), then the probability that he is unable to solve less than two problems is: [April 12, 2019 (II)] (a) \(\frac{201}{5}\left(\frac{1}{5}\right)^{49}\) (b) \(\frac{316}{25}\left(\frac{4}{5}\right)^{48}\) (c) \(\frac{54}{5}\left(\frac{4}{5}\right)^{49}\) (d) \(\frac{164}{25}\left(\frac{1}{5}\right)^{48}\)

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is: \(\quad\) [April 10, 2019 (I)] (a) \(\frac{1}{11}\) (b) \(\frac{1}{10}\) (c) \(\frac{1}{12}\) (d) \(\frac{1}{17}\)

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than \(99 \%\) is : \(\quad\) [April 10, 2019 (II)] (a) 5 (b) 6 (c) 8 (d) 7

Four persons can hit a target correctly with probabilities \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) and \(\frac{1}{8}\) respectively. If all hit at the target independently, then the probability that the target would be hit, is: \(\quad\) [April 09, 2019 (I)] (a) \(\frac{25}{192}\) (b) \(\frac{7}{32}\) (c) \(\frac{1}{192}\) (d) \(\frac{25}{32}\)

Let \(\mathrm{A}\) and \(\mathrm{B}\) be two non-null events such that \(\mathrm{A} \subset \mathrm{B}\). Then, which of the following statements is always correct? [April \(08,2019(\) I) \(]\) (a) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})\) (b) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})\) (c) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \leq \mathrm{P}(\mathrm{A})\) (d) \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=1\)

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least \(90 \%\) is : \(\quad\) [April. 08, 2019 (II)] (a) 5 (b) 3 (c) 4 (d) 2

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to: [Jan. 12,2019 (I)] (a) \(\frac{200}{6^{5}}\) (b) \(\frac{150}{6^{5}}\) (c) \(\frac{225}{6^{5}}\) (d) \(\frac{175}{6^{5}}\)

In a game, a man wins 2100 if he gets 5 or 6 on a throw of a fair die and loses ? 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : \(\quad\) [Jan. 12, 2019 (II)] (a) \(\frac{400}{9}\) loss (b) 0 (c) \(\frac{400}{3}\) gain (d) \(\frac{400}{3}\) loss

Two integers are selected at random from the set \(\\{1,2, \ldots, 11\\}\). Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is : \([\) Jan. \(11,2019(I)]\) (a) \(\frac{7}{10}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{5}\) (d) \(\frac{3}{5}\)

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered \(1,2,3, \ldots, 9\) is randomly picked and the number is èitheer 7 or 8 is: (a) \(\frac{13}{36}\) (b) \(\frac{15}{72}\) (c) \(\frac{19}{72}\) (d) \(\frac{19}{36}\)

If the probability of hitting a target by a shooter, in any shot, is \(\frac{1}{3}\), then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than \(\frac{5}{6}\), is: [Jan. 10, 2019 (II)] (a) 3 (b) 6 (c) 5 (d) 4

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let \(\mathrm{X}\) denote the random variable of number of aces obtained in the two drawn cards. Then \(\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)\) equals: \([\mathrm{Jan} 09, \mathbf{2 0 1 9}(\mathrm{I})]\) (a) \(49 / 169\) (b) \(52 / 169\) (c) \(24 / 169\) (d) \(25 / 169\)

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is: [Jan. 09, 2019 (II)] (a) \(\frac{21}{49}\) (b) \(\frac{27}{49}\) (c) \(\frac{26}{49}\) (d) \(\frac{32}{49}\)

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is : \([2018]\) (a) \(\frac{2}{5}\) (b) \(\frac{1}{5}\) (c) \(\frac{3}{4}\) (d) \(\frac{3}{10}\)

Let \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) be three events, which are pair- wise independence and \(\bar{E}\) denotes the complement of an event E. If \(\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0\) and \(\mathrm{P}(\mathrm{C})>0\), then \(\mathrm{P}[(\bar{A} \cap \bar{B}) \mid \mathrm{C}]\) is equal to. \(\quad\) Online April 16, 2018] (a) \(\mathrm{P}(\mathrm{A})+\mathrm{P}(\bar{B})\) (b) \(\mathrm{P}(\bar{A})-\mathrm{P}(\bar{B})\) (c) \(\mathrm{P}(\bar{A})-\mathrm{P}(\mathrm{B})\) (d) \(\mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})\)

A player \(X\) has a biased coin whose probability of showing heads is \(p\) and a player \(Y\) has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If \(X\) starts the game, and the probability of winning the game by both the players is equal, then the value of ' \(p\) ' is[Online April 15, 2018] (a) \(\frac{1}{3}\) (b) \(\frac{1}{5}\) (c) \(\frac{1}{4}\) (d) \(\frac{2}{5}\)

If two different numbers are taken from the set \((0,1,2,3,\), \(\ldots \ldots, 10\) ), then the probability that their sum as well as absolute difference are both multiple of 4 , is: (a) \(\frac{7}{55}\) (b) \(\frac{6}{55}\) (c) \(\frac{12}{55}\) (d) \(\frac{14}{55}\)

Let \(\mathrm{E}\) and \(\mathrm{F}\) be two independent events. The probability that both \(\mathrm{E}\) and \(\mathrm{F}\) happen is \(\frac{1}{12}\) and the probability that neither E nor F happens is \(\frac{1}{2}\), then a value of \(\frac{\mathrm{P}(\mathrm{E})}{\mathrm{P}(\mathrm{F})}\) is : [Online April 9, 2017] (a) \(\frac{4}{3}\) (b) \(\frac{3}{2}\) (c) \(\frac{1}{3}\) (d) \(\frac{5}{12}\)

Three persons \(\mathrm{P}, \mathrm{Q}\) and \(\mathrm{R}\) independently try to hit a target. If the probabilities of their hitting the target are \(\frac{3}{4}, \frac{1}{2}\) and \(\frac{5}{8}\) respectively, then the probability that the target is hit by P or Q but not by R is : \(\quad\) [Online April 8, 2017] (a) \(\frac{21}{64}\) (b) \(\frac{9}{64}\) (c) \(\frac{15}{64}\) (d) \(\frac{39}{64}\)

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is : [Online April 8, 2017] (a) \(\frac{255}{256}\) (b) \(\frac{127}{128}\) (c) \(\frac{63}{64}\) (d) \(\frac{1}{2}\)

Let two fair six-faced dice \(\mathrm{A}\) and \(\mathrm{B}\) be thrown simultaneously. If \(E_{1}\) is the event that die A shows up four, \(\mathrm{E}_{2}\) is the event that die \(\mathrm{B}\) shows up two and \(\mathrm{E}_{3}\) is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? [2016] (a) \(\mathrm{E}_{1}\) and \(\mathrm{E}_{3}\) are independent. (b) \(\mathrm{E}_{1}, \mathrm{E}_{2}\) and \(\mathrm{E}_{3}\) are independent. (c) \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\) are independent. (d) \(\mathrm{E}_{2}\) and \(\mathrm{E}_{3}\) are independent.

Let \(X\) be a set containing 10 elements and \(P(X)\) be its power set. If A and B are picked up at random from \(\mathrm{P}(\mathrm{X})\), with replacement, then the probability that \(\mathrm{A}\) and \(\mathrm{B}\) have equal number elements, is: \(\quad\) [Online April 10, 2015] (a) \(\frac{\left(2^{10}-1\right)}{2^{10}}\) (b) \(\frac{20^{0} C_{10}}{2^{10}}\) (c) \(\frac{\left(2^{10}-1\right)}{2^{20}}\) (d) \(\frac{20^{0} C_{10}}{2^{20}}\)

Let \(A\) and \(B\) be two events such that \(P(\overline{A \cup B})=\frac{1}{6}\), \(P(\overline{A \cap B})=\frac{1}{4}\) and \(P(\bar{A})=\frac{1}{4}\), where \(\bar{A}\) stands for the complement of the event \(\mathrm{A}\). Then the events \(\mathrm{A}\) and \(\mathrm{B}\) are [2014] (a) independent but not equally likely. (b) independent and equally likely. (c) mutually exclusive and independent. (d) equally likely but not independent.

Let \(\mathrm{A}\) and \(\mathrm{E}\) be any two events with positive probabilities: Statement \(-1: \mathrm{P}(\mathrm{E} / \mathrm{A}) \geq \mathrm{P}(\mathrm{A} / \mathrm{E}) \mathrm{P}(\mathrm{E})\) Statement - 2: \(\mathrm{P}(\mathrm{A} / \mathrm{E}) \geq \mathrm{P}(\mathrm{A} \cap \mathrm{E})\) \mathrm{\\{} O n l i n e ~ A p r i l ~ 1 9 , ~ 2 0 1 4 ] ~ (a) Both the statements are true (b) Both the statements are false (c) Statement- 1 is true, Statement- 2 is false (d) Statement- 1 is false, Statement- 2 is true

A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are \(\frac{3}{4}, \frac{1}{2}, \frac{5}{8} .\) The probability that the target is hit by A or B but not by \(\mathrm{C}\) is : [Online April 23, 2013] (a) \(21 / 64\) (b) \(7 / 8\) (c) \(7 / 32\) (d) \(9 / 64\)

Given two independent events, if the probability that exactly one of them occurs is \(\frac{26}{49}\) and the probability that none of them occurs is \(\frac{15}{49}\), then the probability of more probable of the two events is : [Online April 22, 2013] (a) \(4 / 7\) (b) \(6 / 7\) (c) \(3 / 7\) (d) \(5 / 7\)

The probability of a man hitting a target is \(\frac{2}{5}\). He fires at the target \(k\) times \((k\), a given number). Then the minimum \(k\), so that the probability of hitting the target at least once is more than \(\frac{7}{10}\), is: [Online April9, 2013] (a) 3 (b) 5 (c) 2 (d) 4

Three numbers are chosen at random without replacement from \(\\{1,2,3, . .8\\}\). The probability that their minimum is 3 , given that their maximum is 6, is : [2012] (a) \(\frac{3}{8}\) (b) \(\frac{1}{5}\) (c) \(\frac{1}{4}\) (d) \(\frac{2}{5}\)

Let \(A, B, C\), be pairwise independent events with \(P(\mathrm{C})>0\) and \(P(A \cap B \cap C)=0\). Then \(P\left(A^{c} \cap B^{c} / C\right)\). [2011RS] (a) \(\mathrm{P}\left(B^{c}\right)-\mathrm{P}(B)\) (b) \(P\left(A^{c}\right)+P\left(B^{c}\right)\) (c) \(P\left(A^{c}\right)-P\left(B^{c}\right)\) (d) \(P\left(A^{c}\right)-P(B)\)

If \(C\) and \(D\) are two events such that \(C \subset D\) and \(P(D) \neq 0\), then the correct statement among the following is [2011] (a) \(P(C \mid D) \geq P(C)\) (b) \(P(C \mid D)

One ticket is selected at random from 50 tickets numbered \(00,01,02, \ldots, 49\). Then the probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals: [2009] (a) \(\frac{1}{7}\) (b) \(\frac{5}{14}\) (c) \(\frac{1}{50}\) (d) \(\frac{1}{14}\)

It is given that the events \(A\) and \(B\) are such that \(P(A)=\frac{1}{4}, P(A \mid B)=\frac{1}{2}\) and \(P(B \mid A)=\frac{2}{3} .\) Then \(P(B)\) is [2008] (a) \(\frac{1}{6}\) (b) \(\frac{1}{3}\) (c) \(\frac{2}{3}\) (d) \(\frac{1}{2}\)

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are \(0.3\) and \(0.2\), respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is (a) \(0.2\) (b) \(0.7\) (c) \(0.06\) (d) \(0.14\)

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is (a) \(\frac{2}{9}\) (b) \(\frac{1}{9}\) (c) \(\frac{8}{9}\) (d) \(\frac{7}{9}\)

Let \(A\) and \(B\) be two events such that \(P(\overline{A \cup B})=\frac{1}{6}\), \(P(A \cap B)=\frac{1}{4}\) and \(P(\bar{A})=\frac{1}{4}\), where \(\bar{A}\) stands for complement of event \(A\). Then events \(A\) and \(B\) are [2005] (a) equally likely and mutually exclusive (b) equally likely but not independent (c) independent but not equally likely (d) mutually exclusive and independent

The probability that \(A\) speaks truth is \(\frac{4}{5}\), while the probability for \(B\) is \(\frac{3}{4}\). The probability that they contradict each other when asked to speak on a fact is (a) \(\frac{4}{5}\) (b) \(\frac{1}{5}\) (c) \(\frac{7}{20}\) (d) \(\frac{3}{20}\)

A problem in mathematics is given to three students \(A, B\), \(C\) and their respective probability of solving the problem is \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{4}\). Probability that the problem is solved is \([2002]\) (a) \(\frac{3}{4}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{1}{3}\)

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

In a bombing attack, there is \(50 \%\) chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least \(99 \%\) chance of completely destroying the target, is

The probability of a man hitting a target is \(\frac{1}{10} .\) The least number of shots required, so that the probability of his hitting the target at least once is greater than \(\frac{1}{4}\), is \(\quad\) [NA Sep. 04, 2020 (I)]

Let a random variable X have a binomial distribution with mean 8 and variance 4. If \(\mathrm{P}\left(\mathrm{X} d^{\prime \prime} 2\right)=\frac{k}{2^{16}}\), then \(\mathrm{k}\) is equal to: \([\) April 12, \(2019(\mathrm{I})]\) (a) 17 (b) 121 (c) 1 (d) 137