CBSE Class 12 Mathematics Probability VBQs Set B

Fill in the Blanks

Question. If \( E \) and \( F \) are independent events such that \( P(E) = p \), \( P(F) = 2p \) and \( P(\text{Exactly one of } E, F) = \frac{5}{9} \), then \( p = \) ____________.
Answer: \( \frac{1}{3}, \frac{5}{12} \)

Question. If \( P(A) = \frac{1}{2} \), \( P(A \cap B) = \frac{1}{3} \), then the value of \( P(B/A) = \) ____________.
Answer: \( \frac{2}{3} \)

Question. The mean and variance of a binomial variate \( X \) are 2 and 1 respectively, then value of \( p = \) ____________.
Answer: \( \frac{1}{2} \)

Question. A random variable \( X \) has the following probability distribution:

then the value of \( a = \) ____________.
Answer: \( \frac{1}{81} \)

Very Short Answer Questions

Question. An unbiased coin is tossed 4 times. Find the probability of getting atleast one head.
Answer: When an unbiased coin is tossed once, then \( P(H) = P(T) = \frac{1}{2} \), where \( H \) and \( T \) denote head and tail respectively.
\( \therefore \) Probability of getting atleast one head \( = 1 - P(\text{No head}) \)
\( = 1 - P(\text{all tails}) = 1 - \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \)
\( = 1 - \frac{1}{16} = \frac{15}{16} \)

Question. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let \( A \) be the event "number obtained is even" and \( B \) be the event "number obtained is red". Find if \( A \) and \( B \) are independent events.
Answer: Here, \( A = \) Event that "number obtained is even".
\( B = \) Event that "number obtained is red".
\( P(A) = \frac{3}{6} = \frac{1}{2} \); \( P(B) = \frac{3}{6} = \frac{1}{2} \); \( P(A \cap B) = \frac{1}{6} \)
\( P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
i.e., \( P(A \cap B) \neq P(A) \cdot P(B) \)
Hence, \( A \) and \( B \) are not independent events.

Question. Write the probability of an even prime number on each die, when a pair of dice is rolled.
Answer: The probability of getting even prime number on each die \( = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)
(As there is only one even prime number on each die i.e., 2).

Question. Two Independent events \( A \) and \( B \) are given such that \( P(A) = 0.3 \) and \( P(B) = 0.6 \), find \( P(A \text{ and not } B) \).
Answer: We have, \( P(A \text{ and not } B) = P(A \cap B') = P(A) - P(A \cap B) \)
\( = 0.3 - 0.18 \) [\(\because P(A \cap B) = P(A) \times P(B)\)]
\( = 0.12 \)

Question. The probability distribution of \( X \) is:

Write the value of \( k \).
Answer: We have, \( \sum P(X) = 1 \Rightarrow 0.2 + 4k = 1 \Rightarrow 4k = 0.8 \Rightarrow k = 0.2 \)

Question. The probability that atleast one of the two events \( A \) and \( B \) occurs is 0.6. If \( A \) and \( B \) occur simultaneously with probability 0.3, evaluate \( P(\bar{A}) + P(\bar{B}) \).
Answer: We know that, \( A \cup B \), denotes the occurrence of atleast one of \( A \) and \( B \) and \( A \cap B \) denotes the occurrence of both \( A \) and \( B \), simultaneously.
Thus, \( P(A \cup B) = 0.6 \) and \( P(A \cap B) = 0.3 \)
Also, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow 0.6 = P(A) + P(B) - 0.3 \)
\( \Rightarrow P(A) + P(B) = 0.9 \)
\( \Rightarrow [1 - P(\bar{A})] + [1 - P(\bar{B})] = 0.9 \) [\(\because P(A) = 1 - P(\bar{A}) \text{ and } P(B) = 1 - P(\bar{B})\)]
\( \Rightarrow P(\bar{A}) + P(\bar{B}) = 2 - 0.9 = 1.1 \)

Question. Let \( A \) and \( B \) be two events. If \( P(A) = 0.2, P(B) = 0.4, P(A \cup B) = 0.6 \) then find \( P(A/B) \).
Answer: Since, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( 0.6 = 0.2 + 0.4 - P(A \cap B) \Rightarrow P(A \cap B) = 0.6 - 0.6 = 0 \)
\( \therefore P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{0.4} = 0 \)

Short Answer Questions-I

Question. Prove that if \( E \) and \( F \) are independent events, then the events \( E \) and \( F' \) are also independent.
Answer: Since, \( E \) and \( F \) are independent events.
\( \Rightarrow P(E \cap F) = P(E) \cdot P(F) \)
Now, \( P(E \cap F') = P(E) - P(E \cap F) \)
\( = P(E) - P(E) \cdot P(F) = P(E)(1 - P(F)) \)
\( \Rightarrow P(E \cap F') = P(E) \cdot P(F') \)
Hence, \( E \) and \( F' \) are independent events.

Question. From a set of 100 cards numbered 1 to 100, one card is drawn at random. Find the probability that the number on the card is divisible by 6 or 8, but not by 24.
Answer: Numbers divisible by 6 from 1 to 100 \( = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 \)
Numbers divisible by 8 from 1 to 100 \( = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 \)
\( \therefore \) Numbers divisible by 6 or 8 but not by 24 from 1 to 100 \( = 6, 8, 12, 16, 18, 30, 32, 36, 40, 42, 54, 56, 60, 64, 66, 78, 80, 84, 88, 90. \)
\( \therefore \) Required probability \( = \frac{20}{100} = \frac{1}{5} \).

Question. If \( P(A) = 0.6 \), \( P(B) = 0.5 \) and \( P(B/A) = 0.4 \), find \( P(A \cup B) \) and \( P(A/B) \).
Answer: We have \( P(A) = 0.6 \), \( P(B) = 0.5 \) and \( P(B/A) = 0.4 \)
\( \because P(B/A) = \frac{P(A \cap B)}{P(A)} \Rightarrow P(A \cap B) = P(A) P(B/A) \)
\( \Rightarrow P(A \cap B) = 0.6 \times 0.4 = 0.24 \)
Now \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.6 + 0.5 - 0.24 = 1.1 - 0.24 = 0.86 \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.5} = 0.48 \)

Question. The probability of two students \( A \) and \( B \) coming to school in time are \( \frac{2}{7} \) and \( \frac{4}{7} \), respectively. Assuming that the events ‘\( A \) coming on time’ and ‘\( B \) coming on time’ are independent, find the probability of only one of them coming to school on time.
Answer: \( P(A) = \frac{2}{7} \) i.e.; \( A \) is coming on time, \( P(B) = \frac{4}{7} \) i.e.; \( B \) is coming on time,
\( P(A') = 1 - \frac{2}{7} = \frac{5}{7} \), \( P(B') = 1 - \frac{4}{7} = \frac{3}{7} \)
\( \therefore \) Probability of only one of them coming to school on time \( = P(A)P(B') + P(A')P(B) \)
\( = \frac{2}{7} \times \frac{3}{7} + \frac{5}{7} \times \frac{4}{7} = \frac{6}{49} + \frac{20}{49} = \frac{26}{49} \)

Question. Four cards are drawn one by one with replacement from a well-shuffled deck of playing cards. Find the probability that at least three cards are of diamonds.
Answer: \( X \) be the random variable of drawing at least three cards are of diamonds.
\( P(\text{At least three cards are of diamonds}) \)
\( = P(X = 3) + P(X = 4) \)
\( = \frac{{}^{13}C_3 \times {}^{39}C_1}{{}^{52}C_4} + \frac{{}^{13}C_4}{{}^{52}C_4} \) (This is with replacement, wait, formula should be Binomial Distribution)
Wait, the solution provided in OCR uses combinations \( \frac{1}{{}^{52}C_4} [{}^{13}C_3 \times 39 + {}^{13}C_4] \). Following verbatim solution:
\( = \frac{1}{49 \times 50 \times 51 \times 52 / (4 \times 3 \times 2 \times 1)} \left[ \frac{11 \times 12 \times 13 \times 39}{3 \times 2 \times 1} + \frac{10 \times 11 \times 12 \times 13}{4 \times 3 \times 2 \times 1} \right] \)
\( = \frac{4 \times 3 \times 2 \times 1}{49 \times 50 \times 51 \times 52} \left[ \frac{11 \times 12 \times 13 \times 39 \times 4 + 10 \times 11 \times 12 \times 13}{4 \times 3 \times 2 \times 1} \right] \)
\( = \frac{1}{49 \times 50 \times 51 \times 52} \times 11 \times 12 \times 13 \{156 + 10\} = \frac{11 \times 12 \times 13 \times 166}{49 \times 50 \times 51 \times 52} = 0.04 \)

Question. The mean and variance of a binomial distribution are 12 and 3 respectively. Find the binomial distribution.
Answer: Mean \( = np = 12 \) \( \Rightarrow np = 12 \)
and, variance \( = npq = 3 \) \( \Rightarrow npq = 3 \)
\( \therefore \frac{npq}{np} = \frac{3}{12} = \frac{1}{4} \Rightarrow q = \frac{1}{4} \Rightarrow p = 1 - q = 1 - \frac{1}{4} \)
\( \Rightarrow p = \frac{3}{4} \)
Now, \( np = 12 \Rightarrow n \times \frac{3}{4} = 12 \Rightarrow n = 16 \)
\( \therefore \) Binomial distribution is \( B(16, \frac{3}{4}) \)

Short Answer Questions-II

Question. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl? (ii) atleast one is a girl?
Answer: A family has 2 children, then sample space \( = S = \{BB, BG, GB, GG\} \), where B stands for boy and G for girl.
(i) Let \( A = \) Both are girls \( = \{GG\} \) and \( B = \) The youngest is a girl \( = \{BG, GG\} \)
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{2/4} = \frac{1}{2} \) [\(\because A \cap B = \{GG\}\)]
(ii) Let \( C = \) at least one is a girl \( = \{BG, GB, GG\} \)
Now \( P(A/C) = \frac{P(A \cap C)}{P(C)} = \frac{1/4}{3/4} = \frac{1}{3} \) [\(\because A \cap C = \{GG\}\)]

Question. A committee of 4 students is selected at random from a group consisting of 7 boys and 4 girls. Find the probability that there are exactly 2 boys in the committee, given that at least one girl must be there in the committee.
Answer: Let \( A = \) selection of committee having exactly 2 boys.
\( B = \) selection of committee having at least one girl.
The required probability is \( P(A/B) \). Now, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( P(B) = \frac{{}^4C_1 \times {}^7C_3 + {}^4C_2 \times {}^7C_2 + {}^4C_3 \times {}^7C_1 + {}^4C_4}{{}^{11}C_4} = \frac{\frac{4!}{1!3!} \times \frac{7!}{3!4!} + \frac{4!}{2!2!} \times \frac{7!}{2!5!} + \frac{4!}{3!1!} \times \frac{7!}{1!6!} + \frac{4!}{4!0!}}{\frac{11!}{4!7!}} \)
\( = \frac{4 \times \frac{7 \times 6 \times 5}{3 \times 2} + \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6}{2 \times 1} + 4 \times 7 + 1}{\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}} = \frac{140 + 126 + 28 + 1}{330} = \frac{295}{330} = \frac{59}{66} \)
\( P(A \cap B) = \frac{{}^4C_2 \times {}^7C_2}{{}^{11}C_4} = \frac{\frac{4!}{2!2!} \times \frac{7!}{2!5!}}{\frac{11!}{4!7!}} = \frac{\frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}}{330} = \frac{126}{330} = \frac{21}{55} \)
\( \therefore P(A/B) = \frac{21/55}{59/66} = \frac{21}{55} \times \frac{66}{59} = \frac{126}{295} \)

Question. In a hockey match, both teams \( A \) and \( B \) scored same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the referee was fair or not.
Answer: Let \( E_1, E_2 \) be two events such that
\( E_1 = \) The captain of team ‘\( A \)’ gets a six.
\( E_2 = \) The captain of team ‘\( B \)’ gets a six.
Here, \( P(E_1) = \frac{1}{6} \), \( P(E_2) = \frac{1}{6} \), \( P(E_1') = 1 - \frac{1}{6} = \frac{5}{6} \), \( P(E_2') = 1 - \frac{1}{6} = \frac{5}{6} \)
Now, \( P(\text{winning the match by team } A) = \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \dots \)
\( = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} + \dots = \frac{1/6}{1 - (5/6)^2} = \frac{1/6}{11/36} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11} \)
\( P(\text{winning the match by team } B) = 1 - \frac{6}{11} = \frac{5}{11} \)
Decision: Referee’s decision was fair because team spirit enhances co-operation and unity.

Question. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.
Answer: Let \( E, F \) and \( A \) be three events such that
\( E = \) selection of bag \( A \)
\( F = \) selection of bag \( B \)
\( A = \) getting one red and one black ball out of two
Here, \( P(E) = P(\text{getting 1 or 2 in a throw of die}) = \frac{2}{6} = \frac{1}{3} \)
\( \therefore P(F) = 1 - \frac{1}{3} = \frac{2}{3} \)
Also, \( P(A/E) = P(\text{getting one red and one black if bag A is selected}) = \frac{{}^6C_1 \times {}^4C_1}{{}^{10}C_2} = \frac{24}{45} \)
and \( P(A/F) = P(\text{getting one red and one black if bag B is selected}) = \frac{{}^3C_1 \times {}^7C_1}{{}^{10}C_2} = \frac{21}{45} \)
Now, by theorem of total probability,
\( P(A) = P(E) \cdot P(A/E) + P(F) \cdot P(A/F) \)
\( = \frac{1}{3} \times \frac{24}{45} + \frac{2}{3} \times \frac{21}{45} = \frac{8}{45} + \frac{14}{45} = \frac{22}{45} \)

Question. Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II.
Answer: Let \( E_1 \) be the event of choosing the bag \( I \), \( E_2 \) the event of choosing the bag \( II \) and \( A \) be the event of drawing a red ball.
Then \( P(E_1) = P(E_2) = \frac{1}{2} \)
Also \( P(A/E_1) = P(\text{drawing a red ball from bag I}) = \frac{3}{7} \)
and \( P(A/E_2) = P(\text{drawing a red ball from bag II}) = \frac{5}{11} \)
Now, the probability of drawing a ball from bag \( II \), being given that it is red, is \( P(E_2/A) \). By using Bayes’ theorem, we have
\( P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{35}{68} \)

Question. Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Answer: Let \( E_1, E_2 \) and \( A \) be event such that
\( E_1 = \) Selecting male person
\( E_2 = \) Selecting women (female person)
\( A = \) Selecting grey haired person.
Then \( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2} \)
\( P(A/E_1) = \frac{5}{100}, P(A/E_2) = \frac{0.25}{100} \)
Here, required probability is \( P(E_1/A) \).
\( \therefore P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( \therefore P(E_1/A) = \frac{\frac{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100} + \frac{1}{2} \times \frac{0.25}{100}} = \frac{5}{5 + 0.25} = \frac{5}{5.25} = \frac{500}{525} = \frac{20}{21} \)

Question. Three persons A, B and C apply for a job of manager in a private company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Answer: Let \( E_1, E_2, E_3 \) and \( A \) be events such that
\( E_1 = \) person selected is \( A \); \( E_2 = \) person selected is \( B \); \( E_3 = \) person selected is \( C \)
\( A = \) changes to improve profit does not take place.
Now \( P(E_1) = \frac{1}{7}, P(E_2) = \frac{2}{7}, P(E_3) = \frac{4}{7} \)
\( P(A/E_1) = 1 - \frac{8}{10} = \frac{2}{10} \); \( P(A/E_2) = 1 - \frac{5}{10} = \frac{5}{10} \); \( P(A/E_3) = 1 - \frac{3}{10} = \frac{7}{10} \)
We require \( P(E_3/A) \)
\( P(E_3/A) = \frac{P(E_3) \cdot P(A/E_3)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( = \frac{\frac{4}{7} \times \frac{7}{10}}{\frac{1}{7} \times \frac{2}{10} + \frac{2}{7} \times \frac{5}{10} + \frac{4}{7} \times \frac{7}{10}} = \frac{28}{2 + 10 + 28} = \frac{28}{40} = \frac{7}{10} \)

Question. A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white?
Answer: There may be three situations as events.
\( E_1 = \) bag contains 2 white balls, \( E_2 = \) bag contains 3 white balls,
\( E_3 = \) bag contains all 4 white balls, \( A = \) getting two white balls.
We have required \( P(E_3/A) = ? \)
Now, \( P(E_1) = \frac{1}{3}, P(E_2) = \frac{1}{3}, P(E_3) = \frac{1}{3} \)
\( P(A/E_1) = \frac{{}^2C_2}{{}^4C_2} = \frac{1}{6} \); \( P(A/E_2) = \frac{{}^3C_2}{{}^4C_2} = \frac{3}{6} = \frac{1}{2} \); \( P(A/E_3) = \frac{{}^4C_2}{{}^4C_2} = 1 \)
Now, \( P(E_3/A) = \frac{P(E_3) \cdot P(A/E_3)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1} = \frac{1}{\frac{1}{6} + \frac{1}{2} + 1} = \frac{1}{\frac{1+3+6}{6}} = \frac{1}{10/6} = \frac{6}{10} = \frac{3}{5} \)

Question. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs.
Answer: Let \( A, E_1 \) and \( E_2 \) be the events defined as follows:
\( A : \) cards drawn are both clubs
\( E_1 : \) lost card is club; \( E_2 : \) lost card is not a club
Then, \( P(E_1) = \frac{13}{52} = \frac{1}{4}, P(E_2) = \frac{39}{52} = \frac{3}{4} \)
\( P(A/E_1) = \) Probability of drawing both club cards when lost card is club \( = \frac{12}{51} \times \frac{11}{50} \)
\( P(A/E_2) = \) Probability of drawing both club cards when lost card is not a club \( = \frac{13}{51} \times \frac{12}{50} \)
To find: \( P(E_1/A) \)
By Baye’s Theorem, \( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \)
\( = \frac{\frac{1}{4} \times \frac{12}{51} \times \frac{11}{50}}{\frac{1}{4} \times \frac{12}{51} \times \frac{11}{50} + \frac{3}{4} \times \frac{13}{51} \times \frac{12}{50}} = \frac{12 \times 11}{12 \times 11 + 3 \times 13 \times 12} = \frac{11}{11 + 39} = \frac{11}{50} \)

Question. Often it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is actually a six. Find the probability that it is actually a six.
Answer: Let \( E_1, E_2 \) and \( E \) be three events such that
\( E_1 = \) six occurs; \( E_2 = \) six does not occur and
\( E = \) man reports that six occurs in the throwing of the dice.
Now, \( P(E_1) = \frac{1}{6}, P(E_2) = \frac{5}{6} \); \( P(E/E_1) = \frac{4}{5}, P(E/E_2) = 1 - \frac{4}{5} = \frac{1}{5} \)
We have to find \( P(E_1/E) \)
\( P(E_1/E) = \frac{P(E_1) \cdot P(E/E_1)}{P(E_1) \cdot P(E/E_1) + P(E_2) \cdot P(E/E_2)} \)
\( = \frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}} = \frac{4/30}{4/30 + 5/30} = \frac{4}{9} \)

Question. In shop A, 30 tin pure ghee and 40 tin adulterated ghee are kept for sale while in shop B, 50 tin pure ghee and 60 tin adulterated ghee are there. One tin of ghee is purchased from one of the shops randomly and it is found to be adulterated. Find the probability that it is purchased from shop B.
Answer: Let the event be defined as \( E_1 = \) selection of shop \( A \); \( E_2 = \) selection of shop \( B \) and
\( A = \) purchasing of a tin having adulterated ghee
\( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2}, P(A/E_1) = \frac{40}{70} = \frac{4}{7}, P(A/E_2) = \frac{60}{110} = \frac{6}{11} \)
\( \therefore \) Required probability is given by
\( P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{\frac{1}{2} \cdot \frac{6}{11}}{\frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{6}{11}} = \frac{3/11}{2/7 + 3/11} = \frac{3/11}{43/77} = \frac{21}{43} \)

Question. A random variable \( X \) has the following probability distribution:

Determine: (i) \( k \) (ii) \( P(X < 3) \) (iii) \( P(X > 6) \) (iv) \( P(0 < X < 3) \).
Answer: \( \because \sum p_i = 1 \)
\( \therefore 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1 \)
\( \Rightarrow 10k^2 + 9k - 1 = 0 \)
\( \Rightarrow 10k^2 + 10k - k - 1 = 0 \Rightarrow 10k(k + 1) - 1(k + 1) = 0 \)
\( \Rightarrow (k + 1)(10k - 1) = 0 \Rightarrow k = -1 \text{ and } k = \frac{1}{10} \)
(i) \( \because k \) can never be negative as probability is never negative then \( k = \frac{1}{10} \)
(ii) \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k = \frac{3}{10} \)
(iii) \( P(X > 6) = P(X = 7) = 7k^2 + k = 7 \times \frac{1}{100} + \frac{1}{10} = \frac{17}{100} \)
(iv) \( P(0 < X < 3) = P(X = 1) + P(X = 2) = k + 2k = 3k = \frac{3}{10} \)

Question. Three numbers are selected at random (without replacement) from first six positive integers. If \( X \) denotes the smallest of the three numbers obtained, find the probability distribution of \( X \). Also, find the mean and variance of the distribution.
Answer: First six positive integers are 1, 2, 3, 4, 5 and 6. Number of elements in sample space \( n(S) = {}^6C_3 = 20 \).
X, smallest of the three numbers, can be 1, 2, 3 and 4.
\( P(X = 1) = \frac{{}^5C_2}{20} = \frac{10}{20} = \frac{1}{2} \)
\( P(X = 2) = \frac{{}^4C_2}{20} = \frac{6}{20} = \frac{3}{10} \)
\( P(X = 3) = \frac{{}^3C_2}{20} = \frac{3}{20} \)
\( P(X = 4) = \frac{{}^2C_2}{20} = \frac{1}{20} \)

Mean \( = E(X) = \sum p_i x_i = 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{3}{20} + 4 \cdot \frac{1}{20} = \frac{1}{2} + \frac{6}{10} + \frac{9}{20} + \frac{4}{20} = \frac{10+12+9+4}{20} = \frac{35}{20} = \frac{7}{4} \).
Variance \( = \sum x_i^2 p_i - [E(X)]^2 = \{1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{3}{10} + 3^2 \cdot \frac{3}{20} + 4^2 \cdot \frac{1}{20}\} - (\frac{7}{4})^2 \)
\( = \{\frac{1}{2} + \frac{12}{10} + \frac{27}{20} + \frac{16}{20}\} - \frac{49}{16} = \frac{10+24+27+16}{20} - \frac{49}{16} = \frac{77}{20} - \frac{49}{16} = \frac{308 - 245}{80} = \frac{63}{80} \).

Question. In a game, a man wins ₹5 for getting a number greater than 4 and loses ₹1 otherwise, when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses.
Answer: Let \( X \) be random variable, which is possible value of winning or losing. \( X \) may have value ₹5, ₹4, ₹3 and -₹3.
\( P(X = 5) = P(\text{getting number greater than 4 in first throw}) = \frac{2}{6} = \frac{1}{3} \).
\( P(X = 4) = P(\text{greater than 4 in 2nd throw}) = \frac{4}{6} \times \frac{2}{6} = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \).
\( P(X = 3) = P(\text{greater than 4 in 3rd throw}) = \frac{4}{6} \times \frac{4}{6} \times \frac{2}{6} = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27} \).
\( P(X = -3) = P(\text{greater than 4 in no throw}) = \frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27} \).

Expected value \( = E(X) = \sum x_i p_i = 5 \times \frac{1}{3} + 4 \times \frac{2}{9} + 3 \times \frac{4}{27} + (-3) \times \frac{8}{27} \)
\( = \frac{5}{3} + \frac{8}{9} + \frac{12}{27} - \frac{24}{27} = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \frac{19}{9} = Rs 2 \frac{1}{9} \).

Question. There are 4 cards numbered 1 to 4, one number on one card. Two cards are drawn at random without replacement. Let \( X \) denote the sum of the numbers on the two drawn cards. Find the mean and variance of \( X \).
Answer: Sample space \( S = \{(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (4,1), (4,2), (4,3), (3,1), (3,2), (3,4)\} \). \( n(S) = 12 \).
\( X \), sum of numbers, can be 3, 4, 5, 6, 7.
\( P(X = 3) = P\{(1,2), (2,1)\} = \frac{2}{12} = \frac{1}{6} \).
\( P(X = 4) = P\{(1,3), (3,1)\} = \frac{2}{12} = \frac{1}{6} \).
\( P(X = 5) = P\{(1,4), (4,1), (2,3), (3,2)\} = \frac{4}{12} = \frac{1}{3} \).
\( P(X = 6) = P\{(4,2), (2,4)\} = \frac{2}{12} = \frac{1}{6} \).
\( P(X = 7) = P\{(4,3), (3,4)\} = \frac{2}{12} = \frac{1}{6} \).

Mean \( = \sum X.P(X) = \frac{3}{6} + \frac{4}{6} + \frac{5}{3} + \frac{6}{6} + \frac{7}{6} = \frac{3+4+10+6+7}{6} = \frac{30}{6} = 5 \).
Variance \( = \sum X^2 P(X) - (\sum X.P(X))^2 = \{\frac{9}{6} + \frac{16}{6} + \frac{25}{3} + \frac{36}{6} + \frac{49}{6}\} - (5)^2 \)
\( = \frac{9 + 16 + 50 + 36 + 49}{6} - 25 = \frac{160}{6} - 25 = \frac{160 - 150}{6} = \frac{10}{6} = \frac{5}{3} \).

Question. The random variable \( X \) can take only the values 0, 1, 2, 3. Given that \( P(X = 0) = P(X = 1) = p \) and \( P(X = 2) = P(X = 3) = a \) such that \( \sum p_i x_i^2 = 2 \sum p_i x_i \), find the value of \( p \).
Answer: Given \( X \) is a random variable with values 0, 1, 2, 3.

\( \therefore \sum x_i p_i = 0 + p + 2a + 3a = p + 5a \)
\( \sum x_i^2 p_i = 0 + p + 4a + 9a = p + 13a \)
According to question:
\( p + 13a = 2(p + 5a) \Rightarrow p + 13a = 2p + 10a \Rightarrow p = 3a \)
Also \( p + p + a + a = 1 \Rightarrow 2p + 2a = 1 \)
\( 2a = 1 - 2p \Rightarrow a = \frac{1 - 2p}{2} \)
\( \therefore p = 3 \times \frac{1 - 2p}{2} \Rightarrow 2p = 3 - 6p \Rightarrow 8p = 3 \Rightarrow p = \frac{3}{8} \).