CBSE Class 12 Mathematics Probability Worksheet Set 01

Very Short Answer Questions

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., \( P(A \cap B) \neq P(A).P(B) \)
Hence, \(A\) and \(B\) are not independent events.

Question. If \( P(A) = \frac{1}{2}, P(B) = 0 \), then find \( P(A / B) \).
Answer: We have, \( P(A) = \frac{1}{2}, P(B) = 0 \)
\( \therefore P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Since, \( P(B) = 0 \), so \( P(A/B) \) is not defined.

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 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,
\( P(A \text{ and not } B) = P(A \cap B') = P(A) - P(A \cap B) \)
\( = 0.3 - 0.18 \quad [\because P(A \cap B) = P(A) \times P(B)] \)
\( = 0.12 \)

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

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

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 \( P(\overline{A}) + P(\overline{B}) \). 
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, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \Rightarrow \) \( 0.6 = P(A) + P(B) - 0.3 \)
\( \Rightarrow \) \( P(A) + P(B) = 0.9 \)
\( \Rightarrow \) \( [1 - P(\overline{A})] + [1 - P(\overline{B})] = 0.9 \quad [\because P(A) = 1 - P(\overline{A}) \text{ and } P(B) = 1 - P(\overline{B})] \)
\( \Rightarrow \) \( P(\overline{A}) + P(\overline{B}) = 2 - 0.9 = 1.1 \)

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 \( P\left(\frac{A}{B}\right) \). 
Answer: Since, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( 0.6 = 0.2 + 0.4 - P(A \cap B) \)
\( \Rightarrow \) \( P(A \cap B) = 0.6 - 0.6 = 0 \)
\( \therefore P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} = \frac{0}{0.4} = 0 \)
 

Short Answer Questions 

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.
\( \Rightarrow \) \( P(E \cap F) = P(E) . P(F) \)
Now, \( P(E \cap F') = P(E) - P(E \cap F) \)
\( = P(E) - P(E) . P(F) = P(E)(1 - P(F)) \)
\( \Rightarrow \) \( P(E \cap F') = P(E) . P(F') \)
Hence, \( E \) and \( F' \) are independent events.

Question. If \( P(A) = 0.4 \), \( P(B) = p \), \( P(A \cup B) = 0.6 \) and \(A\) and \(B\) are given to be independent events, find the value of ‘\(p\)’. 
Answer: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( \Rightarrow \) \( 0.6 = 0.4 + p - P(A \cap B) \)
\( \Rightarrow \) \( P(A \cap B) = 0.4 + p - 0.6 = p - 0.2 \)
Since, \(A\) and \(B\) are independent events.
\( \therefore P(A \cap B) = P(A) \times P(B) \)
\( \Rightarrow \) \( p - 0.2 = 0.4 \times p \)
\( \Rightarrow \) \( p - 0.4p = 0.2 \)
\( \Rightarrow \) \( 0.6p = 0.2 \)
\( \Rightarrow \) \( p = \frac{0.2}{0.6} = \frac{1}{3} \)

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: Number divisible by 6 from 1 to 100 \( = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 \)
Number divisible by 8 from 1 to 100 \( = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 \)
\( \therefore \) Number divisible by 6 or 8 but not by 24 from 1 to 100 \( = 6, 8, 12, 16, 18, 30, 32, 36, 40, 42, 54, 56, 60, 64, 66, 78, 80, 84, 88, 90 \).
\( \therefore \text{Required probability} = \frac{20}{100} = \frac{1}{5} \).

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 \( B \) comming to school in time are \( \frac{2}{7} \) and \( \frac{4}{7} \), respectively. Assuming that the events ‘\(A\) comming on time’ and ‘\(B\) comming on time’ are independent, find the probability of only one of them comming to school on time. 
Answer: \( P(A) = \frac{2}{7} \) i.e.; \( A \) is comming on time, \( P(B) = \frac{4}{7} \) i.e.; \( B \) is comming 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 comming 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: Let \( X \) be the random variable of drawing at least three cards are of diamonds.
\( P(\text{At least three cards are of diamonds}) \)
\( = P(X = 3) + P(X = 4) \)
\( = \frac{^{13}C_3 \times ^{39}C_1}{^{52}C_4} + \frac{^{13}C_4}{^{52}C_4} = \frac{1}{^{52}C_4}[^{13}C_3 \times 39 + ^{13}C_4] \)
\( = \frac{1}{\frac{52!}{4! 48!}} \left[ \frac{13!}{3! 10!} \times 39 + \frac{13!}{4! 9!} \right] \)
\( = \frac{1}{\frac{49 \times 50 \times 51 \times 52}{4 \times 3 \times 2 \times 1}} \left[ \frac{11 \times 12 \times 13 \times 39}{3 \times 2 \times 1} + \frac{10 \times 11 \times 12 \times 13}{4 \times 3 \times 2 \times 1} \right] \)
\( = \frac{4 \times 3 \times 2 \times 1}{49 \times 50 \times 51 \times 52} \left[ \frac{11 \times 12 \times 13 \times 39 \times 4 + 10 \times 11 \times 12 \times 13}{4 \times 3 \times 2 \times 1} \right] \)
\( = \frac{1}{49 \times 50 \times 51 \times 52} \times 11 \times 12 \times 13 \{156 + 10\} = \frac{11 \times 12 \times 13 \times 166}{49 \times 50 \times 51 \times 52} = 0.04 \)

Long Answer Questions-I

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: \( \because \sum_{i=1}^n P_i = 1 \)
\( \therefore 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1 \)
\( \Rightarrow \) \( 10k^2 + 9k - 1 = 0 \)
\( \Rightarrow \) \( 10k^2 + 10k - k - 1 = 0 \)
\( \Rightarrow \) \( 10k(k + 1) - 1(k + 1) = 0 \)
\( \Rightarrow \) \( (k + 1)(10k - 1) = 0 \)
\( \Rightarrow \) \( k = -1 \text{ and } k = \frac{1}{10} \)
(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.
If three numbers are selected at random from above six numbers then the number of elements in sample space S is given by
i.e., \( n(S) = ^6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2} = 20 \)
Here \( X \), smallest of the three numbers obtained, is random variable \( X \) may have value 1, 2, 3 and 4. Therefore, required probability distribution is given as
\( P(X = 1) = \) Probability of event getting 1 as smallest number
\( = \frac{^5C_2}{20} = \frac{5!}{2! 3! \times 20} = \frac{5 \times 4}{2 \times 20} = \frac{10}{20} = \frac{1}{2} \quad [^5C_2 = \text{selection of two numbers out of } 2, 3, 4, 5, 6] \)
\( P(X = 2) = \) Probability of events getting 2 as smallest number.
\( = \frac{^4C_2}{20} = \frac{4!}{2! 2! \times 20} = \frac{6}{20} = \frac{3}{10} \quad [^4C_2 = \text{selection of two numbers out of } 3, 4, 5, 6] \)
\( P(X = 3) = \) Probability of events getting 3 as smallest number
\( = \frac{^3C_2}{20} = \frac{3!}{2! 1! \times 20} = \frac{3}{20} \quad [^3C_2 = \text{selection of two numbers out of } 4, 5, 6] \)
\( P(X = 4) = \) Probability of events getting 4 as smallest number.
\( = \frac{^2C_2}{20} = \frac{1}{20} \quad [^2C_2 = \text{selection of two numbers out of } 5, 6] \)
Required probability distribution table is

Mean \( = E(X) = \Sigma p_i x_i \)
\( = 1 \times \frac{1}{2} + 2 \times \frac{3}{10} + 3 \times \frac{3}{20} + 4 \times \frac{1}{20} \)
\( = \frac{1}{2} + \frac{6}{10} + \frac{9}{20} + \frac{4}{20} = \frac{10 + 12 + 9 + 4}{20} = \frac{35}{20} = \frac{7}{4} \)
Variance \( = \Sigma x_i^2 p_i - (EX)^2 \)
\( = \left\{ 1^2 \times \frac{1}{2} + 2^2 \times \frac{3}{10} + 3^2 \times \frac{3}{20} + 4^2 \times \frac{1}{20} \right\} - \left( \frac{7}{4} \right)^2 \)
\( = \frac{1}{2} + \frac{12}{10} + \frac{27}{20} + \frac{16}{20} - \frac{49}{16} = \frac{10 + 24 + 27 + 16}{20} - \frac{49}{16} \)
\( = \frac{77}{20} - \frac{49}{16} = \frac{308 - 245}{80} = \frac{63}{80} \)

Question. 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 \( A \) and \( B \) be two event such that
\( A = \text{Both are girls} = \{GG\} \quad \text{and} \quad B = \text{The youngest is a girl} = \{BG, GG\} \)
\( P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{2}{4}} = \frac{1}{2} \quad [\because A \cap B = \{GG\}] \)
(ii) Let \( C \) be event such that
\( C = \text{at least one is a girl} = \{BG, GB, GG\} \)
Now \( P(A/C) = \frac{P(A \cap C)}{P(C)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \quad [\because A \cap C = \{GG\}] \)

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. Three numbers are selected at random (without replacement) from first six positive integers. Let \( X \) denote the largest 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, 6.
If three numbers are selected at random from above six numbers then the sample space S have 20 elements as
\( n(S) = ^6C_3 = \frac{6!}{3! 3!} = \frac{6 \times 5 \times 4}{3 \times 2} = 20 \)
Here \( X \), greatest of the three numbers obtained, is random variable. X may have value 3, 4, 5, 6. Therefore required probability distribution is given as
\( P(X = 3) = \) Probability of event getting 3 as greatest number.
\( = \frac{^2C_2}{20} = \frac{1}{20} \quad [^2C_2 = \text{selection of two numbers out of } 1, 2] \)
\( P(X = 4) = \) Probability of event getting 4 as greatest number.
\( = \frac{^3C_2}{20} = \frac{3!}{2! 1! \times 20} = \frac{3}{20} \quad [^3C_2 = \text{selection of two numbers out of } 1, 2, 3] \)
\( P(X = 5) = \) Probability of event getting 5 as greatest number.
\( = \frac{^4C_2}{20} = \frac{4!}{2! 2! \times 20} = \frac{6}{20} \quad [^4C_2 = \text{selection of two numbers out of } 1, 2, 3, 4] \)
\( P(X = 6) = \) Probability of event getting 6 as greatest number.
\( = \frac{^5C_2}{20} = \frac{5!}{2! 3! 20} = \frac{10}{20} \quad [^5C_2 = \text{selection of two numbers out of } 1, 2, 3, 4, 5] \)
Required probability distribution in tabular form as

Mean \( = E(X) = \Sigma p_i x_i = 3 \times \frac{1}{20} + 4 \times \frac{3}{20} + 5 \times \frac{6}{20} + 6 \times \frac{10}{20} \)
\( = \frac{3}{20} + \frac{12}{20} + \frac{30}{20} + \frac{60}{20} = \frac{105}{20} = \frac{21}{4} = 5.25 \)
Variance \( = \Sigma p_i x_i^2 - (E(X))^2 \)
\( = \left\{ 3^2 \times \frac{1}{20} + 4^2 \times \frac{3}{20} + 5^2 \times \frac{6}{20} + 6^2 \times \frac{10}{20} \right\} - \left(\frac{21}{4}\right)^2 \)
\( = \left( \frac{9}{20} + \frac{48}{20} + \frac{150}{20} + \frac{360}{20} \right) - \frac{441}{16} = \frac{567}{20} - \frac{441}{16} \)
\( = 28.35 - 27.56 = 0.79 \)

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 \( P(E_1) = \frac{1}{2} \), \( P(E_2) = \frac{1}{2} \)
\( P\left(\frac{A}{E_1}\right) = \frac{5}{100} \), \( P\left(\frac{A}{E_2}\right) = \frac{0.25}{100} \)
Here, required probability is \( P\left(\frac{E_1}{A}\right) \).
\( \therefore P\left(\frac{E_1}{A}\right) = \frac{P(E_1) . P\left(\frac{A}{E_1}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right)} \)
\( \therefore P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100} + \frac{1}{2} \times \frac{0.25}{100}} = \frac{5}{5 + 0.25} = \frac{500}{525} = \frac{20}{21} \)

Question. Suppose 10000 tickets are sold in a lottery each for Rs. 1. First prize is of Rs. 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, then what is your expectation? 
Answer: Let X is the random variable for the prize.

Since, \( E(X) = \Sigma X P(X) \)
\( = 0 \times \frac{9995}{10000} + 500 \times \frac{3}{10000} + 2000 \times \frac{1}{10000} + 3000 \times \frac{1}{10000} \)
\( = \frac{1500 + 2000 + 3000}{10000} = \frac{6500}{10000} = \frac{13}{20} = \text{Rs. } 0.65 \)

Question. The probability that \( A \) hits a target is \( \frac{1}{3} \) and the probability that \( B \) hits it is \( \frac{2}{5} \) If each one of \( A \) and \( B \) shoots at the target, what is the probability that
(i) the target is hit? (ii) exactly one of them hits the target? 
Answer: Let \( P(A) = \) Probability that \( A \) hits the target \( = \frac{1}{3} \)
\( P(B) = \) Probability that \( B \) hits the target \( = 2/5 \)
(i) P(target is hit) = P(at least one of A, B hits)
= 1 - P (none hits)
\( = 1 - \frac{2}{3} \times \frac{3}{5} = \frac{9}{15} = \frac{3}{5} \)
(ii) P(exactly one of them hits) \( = P (A \text{ and } \overline{B} \text{ or } \overline{A} \text{ and } B) = P(A \cap \overline{B} \cup \overline{A} \cap B) \)
\( = P(A) \times P(\overline{B}) + P(\overline{A}) \times P(B) = \frac{1}{3} \times \frac{3}{5} + \frac{2}{3} \times \frac{2}{5} = \frac{7}{15} \)

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 \( A \) and \( F = \) 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, \( P(A/E) = P (\text{getting one red and one black if bag } A \text{ is selected}) = \frac{^6C_1 \times ^4C_1}{^{10}C_2} = \frac{24}{45} \)
and \( P(A/F) = P (\text{getting one red and one black if bag } B \text{ is selected}) = \frac{^3C_1 \times ^7C_1}{^{10}C_2} = \frac{21}{45} \)
Now, by theorem of total probability,
\( P(A) = P (E) . P (A/E) + P (F) . P (A/F) \)
\( = \frac{1}{3} \times \frac{24}{45} + \frac{2}{3} \times \frac{21}{45} = \frac{8 + 14}{45} = \frac{22}{45} \)

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
\( E_1 = \text{person selected is } A \); \( E_2 = \text{person selected is } B \); \( E_3 = \text{person selected is } C \)
\( A = \text{changes to improve profit does not take place.} \)
Now \( P(E_1) = \frac{1}{7}, P(E_2) = \frac{2}{7}, P(E_3) = \frac{4}{7} \)
\( P\left(\frac{A}{E_1}\right) = 1 - \frac{8}{10} = \frac{2}{10} \); \( P\left(\frac{A}{E_2}\right) = 1 - \frac{5}{10} = \frac{5}{10} \); \( P\left(\frac{A}{E_3}\right) = 1 - \frac{3}{10} = \frac{7}{10} \)
We require \( P\left(\frac{E_3}{A}\right) \)
\( P\left(\frac{E_3}{A}\right) = \frac{P(E_3) . P\left(\frac{A}{E_3}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right) + P(E_3) . P\left(\frac{A}{E_3}\right)} = \frac{\frac{4}{7} \times \frac{7}{10}}{\frac{1}{7} \times \frac{2}{10} + \frac{2}{7} \times \frac{5}{10} + \frac{4}{7} \times \frac{7}{10}} \)
\( = \frac{28}{70} \times \frac{70}{2 + 10 + 28} = \frac{28}{40} = \frac{7}{10} \)

Question. In a game, a man wins Rs. 5 for getting a number greater than 4 and loses Rs. 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 loosing of rupee occur with probability of getting a number greater than 4 in 1st, 2nd, 3rd or in any throw respectively.
Obviously X may have value Rs. 5, Rs. 4, Rs. 3 and – Rs. 3 respectively.
Now, \( P(X = 5) = P (\text{getting number greater than 4 in first throw}) \)
\( = \frac{2}{6} = \frac{1}{3} \)
\( P(X = 4) = P (\text{getting number greater than 4 in 2nd throw}) \)
\( = \frac{4}{6} \times \frac{2}{6} = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \)
\( P(X = 3) = P (\text{getting number greater than 4 in 3rd throw}) \)
\( = \frac{4}{6} \times \frac{4}{6} \times \frac{2}{6} = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27} \)
\( P(X = -3) = P (\text{getting number greater than 4 in no throw}) \)
\( = \frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27} \)
Therefore, probability distribution is as

\( \therefore \text{Expected value of the amount he wins/loses} = E(X) \)
\( E(X) = \Sigma x_i p_i = 5 \times \frac{1}{3} + 4 \times \frac{2}{9} + 3 \times \frac{4}{27} + (-3) \times \frac{8}{27} \)
\( = \frac{5}{3} + \frac{8}{9} + \frac{12}{27} - \frac{24}{27} = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \text{Rs. } \frac{19}{9} = \text{Rs. } 2\frac{1}{9} \)

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.
\( E_1 = \text{bag contains 2 white balls}, \quad E_2 = \text{bag contains 3 white balls}, \)
\( E_3 = \text{Bag contains all 4 white balls}, \quad A = \text{Getting two white balls}. \)
We have required \( P\left(\frac{E_3}{A}\right) = ? \)
Now, \( P(E_1) = \frac{1}{3}, \quad P(E_2) = \frac{1}{3}, \quad P(E_3) = \frac{1}{3} \)
\( P\left(\frac{A}{E_1}\right) = \frac{^2C_2}{^4C_2} = \frac{1}{6}; \quad P\left(\frac{A}{E_2}\right) = \frac{^3C_2}{^4C_2} = \frac{3}{6} = \frac{1}{2}; \quad P\left(\frac{A}{E_3}\right) = \frac{^4C_2}{^4C_2} = 1 \)
Now, \( P\left(\frac{E_3}{A}\right) = \frac{P(E_3) . P\left(\frac{A}{E_3}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right) + P(E_3) . P\left(\frac{A}{E_3}\right)} \)
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1} = \frac{\frac{1}{3}}{\frac{1}{18} + \frac{1}{6} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{10}{18}} = \frac{1}{3} \times \frac{18}{10} = \frac{3}{5} \)

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, then that at least one girl must be there in the committee. 
Answer: Let \( A \) and \( B \) be two events such that
\( A = \text{selection of committee having exactly 2 boys.} \)
\( B = \text{selection of committee having at least one girl.} \)
The required probability is \( P\left(\frac{A}{B}\right) \)
Now, \( P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} \)
\( P(B) = \frac{^4C_1 \times ^7C_3 + ^4C_2 \times ^7C_2 + ^4C_3 \times ^7C_1 + ^4C_4}{^{11}C_4} = \frac{4 \times \frac{7!}{3!4!} + \frac{4!}{2!2!} \times \frac{7!}{2!5!} + \frac{4!}{3!1!} \times \frac{7!}{1!6!} + \frac{4!}{4!0!}}{\frac{11!}{4!7!}} \)
\( = \frac{4 \times \frac{7 \times 6 \times 5}{3 \times 2} + \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6}{2 \times 1} + 4 \times 7 + 1}{\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}} = \frac{140 + 126 + 28 + 1}{330} = \frac{295}{330} = \frac{59}{66} \)
\( P(A \cap B) = \frac{^4C_2 \times ^7C_2}{^{11}C_4} = \frac{\frac{4!}{2!2!} \times \frac{7!}{2!5!}}{\frac{11!}{4!7!}} = \frac{\frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}}{330} = \frac{126}{330} = \frac{21}{55} \)
\( \therefore P\left(\frac{A}{B}\right) = \frac{\frac{21}{55}}{\frac{59}{66}} = \frac{21}{55} \times \frac{66}{59} = \frac{126}{295} \)

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: If two cards, from four cards having numbers 1, 2, 3, 4 each are drawn at random then sample space \( S \) is given by
\( S = \{(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (4, 1), (4, 2), (4, 3), (3, 1), (3, 2), (3, 4)\} \)
Let \( X \), sum of the numbers, be random variable. \( X \) may have values 3, 4, 5, 6, 7.
Now \( P(X = 3) = \text{Probability of event getting } (1, 2), (2, 1) = \frac{2}{12} = \frac{1}{6} \)
\( P(X = 4) = \text{Probability of event getting } (1, 3), (3, 1) = \frac{2}{12} = \frac{1}{6} \)
\( P(X = 5) = \text{Probability of event getting } (1, 4), (4, 1), (2, 3), (3, 2) = \frac{4}{12} = \frac{1}{3} \)
\( P(X = 6) = \text{Probability of event getting } (4, 2), (2, 4) = \frac{2}{12} = \frac{1}{6} \)
\( P(X = 7) = \text{Probability of event getting } (4, 3), (3, 4) = \frac{2}{12} = \frac{1}{6} \)
Thus, probability distribution is represented in tabular form as

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

Question. 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, \( P(E_1) = \frac{13}{52} = \frac{1}{4} \), \( P(E_2) = \frac{39}{52} = \frac{3}{4} \)
\( P(A/E_1) = \text{Probability of drawing both club cards when lost card is club} = \frac{12}{51} \times \frac{11}{50} \)
\( P(A/E_2) = \text{Probability of drawing both club cards when lost card is not a club} = \frac{13}{51} \times \frac{12}{50} \)
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. The random variable \( X \) can take only the values 0, 1, 2, 3. Given that \( P(X = 0) = P(X = 1) = p \) and \( P(X = 2) = P(X = 3) \) such that \( \Sigma p_i x_i^2 = 2 \Sigma p_i x_i \), find the value of \( p \). 
Answer: Given \( X \) is a random variable with values 0, 1, 2, 3. Given probability distributions are as

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

Question. Three machines \( E_1, E_2, E_3 \) in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric tubes. It is known that 4% of the tube produced on each of machines \( E_1 \) and \( E_2 \) are defective and that 5% of those produced on \( E_3 \), are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective. 
Answer: Let \( A \) be the event that the picked up tube is defective.
Let \( A_1, A_2, A_3 \) be events such that
\( A_1 = \text{event of producing tube by machine } E_1 \)
\( A_2 = \text{event of producing tube by machine } E_2 \)
\( A_3 = \text{event of producing tube by machine } E_3 \)
\( P(A_1) = \frac{50}{100} = \frac{1}{2}, \quad P(A_2) = \frac{25}{100} = \frac{1}{4}, \quad P(A_3) = \frac{25}{100} = \frac{1}{4} \)
Also, \( P\left(\frac{A}{A_1}\right) = \frac{4}{100} = \frac{1}{25}; \quad P\left(\frac{A}{A_2}\right) = \frac{4}{100} = \frac{1}{25} \quad \text{and} \quad P\left(\frac{A}{A_3}\right) = \frac{5}{100} = \frac{1}{20} \)
Now, \( P(A) \) is required.
From concept of total probability,
\( P(A) = P(A_1) . P\left(\frac{A}{A_1}\right) + P(A_2) . P\left(\frac{A}{A_2}\right) + P(A_3) . P\left(\frac{A}{A_3}\right) \)
\( = \frac{1}{2} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{20} = \frac{1}{50} + \frac{1}{100} + \frac{1}{80} \)
\( = \frac{8 + 4 + 5}{400} = \frac{17}{400} = 0.0425 \)

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
\( E_1 = \text{six occurs; } E_2 = \text{six does not occur and} \)
\( E = \text{man reports that six occurs in the throwing of the dice.} \)
Now, \( P(E_1) = \frac{1}{6}, P(E_2) = \frac{5}{6}; \quad P\left(\frac{E}{E_1}\right) = \frac{4}{5}, P\left(\frac{E}{E_2}\right) = 1 - \frac{4}{5} = \frac{1}{5} \)
We have to find \( P\left(\frac{E_1}{E}\right) \)
\( P\left(\frac{E_1}{E}\right) = \frac{P(E_1) . P\left(\frac{E}{E_1}\right)}{P(E_1) . P\left(\frac{E}{E_1}\right) + P(E_2) . P\left(\frac{E}{E_2}\right)} = \frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}} = \frac{4}{30} \times \frac{30}{4 + 5} = \frac{4}{9} \)

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
\( E_1 = \text{selection of shop A; } E_2 = \text{selection of shop B and} \)
\( A = \text{purchasing of a tin having adulterated ghee} \)
\( P(E_1) = \frac{1}{2}, \quad P(E_2) = \frac{1}{2}, \quad P\left(\frac{A}{E_1}\right) = \frac{40}{70} = \frac{4}{7}, \quad P\left(\frac{A}{E_2}\right) = \frac{60}{110} = \frac{6}{11} \)
\( \therefore \text{Required probability is given by} \)
\( P\left(\frac{E_2}{A}\right) = \frac{P(E_2) . P\left(\frac{A}{E_2}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right)} = \frac{\frac{1}{2} \cdot \frac{6}{11}}{\frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{6}{11}} = \frac{\frac{3}{11}}{\frac{2}{7} + \frac{3}{11}} = \frac{21}{43} \)

Question. The probabilities of two students \( A \) and \( B \) coming to the school in time are \( \frac{3}{7} \) and \( \frac{5}{7} \) respectively. Assuming that the events, ‘\( A \) coming in time’ and ‘\( B \) coming in time’ are independent, find the probability of only one of them coming to the school in time. 
Answer: Let \( E_1 \) and \( E_2 \) be two events such that
\( E_1 = A \text{ coming to the school in time}, \quad E_2 = B \text{ coming to the school in time.} \)
Here, \( P(E_1) = \frac{3}{7} \text{ and } P(E_2) = \frac{5}{7}; \quad P(\overline{E}_1) = \frac{4}{7}, P(\overline{E}_2) = \frac{2}{7} \)
P (only one of them coming to the school in time) \( = P(E_1) \times P(\overline{E}_2) + P(\overline{E}_1) \times P(E_2) \)
\( = \frac{3}{7} \times \frac{2}{7} + \frac{5}{7} \times \frac{4}{7} \)
\( = \frac{6}{49} + \frac{20}{49} = \frac{26}{49} \)

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
\( E_1 = \text{The captain of team ‘}A\text{’ gets a six.} \)
\( E_2 = \text{The captain of team ‘}B\text{’ gets a six.} \)
Here, \( P(E_1) = \frac{1}{6}, \quad P(E_2) = \frac{1}{6} \quad P(E_1') = 1 - \frac{1}{6} = \frac{5}{6}, P(E_2') = 1 - \frac{1}{6} = \frac{5}{6} \)
Now, \( P \text{ (winning the match by team A)} = \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + ... \)
\( = \frac{1}{6} + \left(\frac{5}{6}\right)^2 . \frac{1}{6} + \left(\frac{5}{6}\right)^4 . \frac{1}{6} + ...... = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11} \)
\( P \text{ (winning the match by team B)} = 1 - \frac{6}{11} = \frac{5}{11} \)
[Note: If \( a \) be the first term and \( r \) the common ratio then sum of infinite terms \( S_\infty = \frac{a}{1 - r} \)]

Question. A Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution. 
Answer: Let \( X \) be the number of bad oranges in two draws of orange from the lot. Here, \( X \) is random variable and may have value 0, 1, 2.
Now, \( P(X = 0) = \frac{^{16}C_2}{^{20}C_2} = \frac{16!}{2! \times 14!} \times \frac{2! \times 18!}{20!} = \frac{18 \times 17 \times 16 \times 15}{20 \times 19 \times 18 \times 17} = \frac{60}{95} \)
\( P(X = 1) = \frac{^4C_1 \times ^{16}C_1}{^{20}C_2} = \frac{4 \times 16 \times 2}{20 \times 19} = \frac{32}{95}, \quad P(X = 2) = \frac{^4C_2}{^{20}C_2} = \frac{4 \times 3}{20 \times 19} = \frac{3}{95} \)
Now, required probability distribution is

Now, Mean \( = \Sigma X_i P(X_i) = 0 \times \frac{60}{95} + 1 \times \frac{32}{95} + 2 \times \frac{3}{95} = \frac{32}{95} + \frac{6}{95} = \frac{38}{95} = \frac{2}{5} \)
Variance \( = \Sigma X_i^2 P(X_i) - \Sigma X_i P(X_i)^2 = 1 \times \frac{32}{95} + 4 \times \frac{3}{95} - \left(\frac{2}{5}\right)^2 = \frac{44}{95} - \frac{4}{25} = \frac{144}{475} \)

Long Answer Questions-II

Question. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade. 
Answer: Let \( E_1, E_2, E_3, E_4 \) and \( A \) be event defined as
\( E_1 = \text{the lost card is a spade card, } E_2 = \text{the lost card is a non spade card} \)
and \( A = \text{drawing three spade cards from the remaining cards.} \)
Now, \( P(E_1) = \frac{13}{52} = \frac{1}{4}, P(E_2) = \frac{39}{52} = \frac{3}{4} \)
\( P\left(\frac{A}{E_1}\right) = \frac{^{12}C_3}{^{51}C_3} = \frac{220}{20825}; \quad P\left(\frac{A}{E_2}\right) = \frac{^{13}C_3}{^{51}C_3} = \frac{286}{20825} \)
Here, required probability = \( P\left(\frac{E_1}{A}\right) \)
\( \therefore P\left(\frac{E_1}{A}\right) = \frac{P(E_1) . P\left(\frac{A}{E_1}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right)} = \frac{\frac{1}{4} \times \frac{220}{20825}}{\frac{1}{4} \times \frac{220}{20825} + \frac{3}{4} \times \frac{286}{20825}} \)
\( = \frac{220}{220 + 3 \times 286} = \frac{220}{1078} = \frac{10}{49} \)

Question. Bag \( I \) contains 3 red and 4 black balls and bag \( II \) contains 4 red and 5 black balls. Two balls are transferred at random from bag \( I \) to bag \( II \) and then a ball is drawn from bag \( II \). The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black. 
Answer: Let \( E_1, E_2, E_3 \) and \( A \) be events such that
\( E_1 = \text{both transferred balls from bag I to bag II are red.} \)
\( E_2 = \text{both transferred balls from bag I to bag II are black.} \)
\( E_3 = \text{out of two transferred balls one is red and other is black.} \)
\( A = \text{drawing a red ball from bag II.} \)
Here, \( P\left(\frac{E_2}{A}\right) \) is required.
Now, \( P(E_1) = \frac{^3C_2}{^7C_2} = \frac{3!}{2! 1!} \times \frac{2! 5!}{7!} = \frac{1}{7}; \quad P(E_2) = \frac{^4C_2}{^7C_2} = \frac{4!}{2! 2!} \times \frac{2! 5!}{7!} = \frac{2}{7} \)
\( P(E_3) = \frac{^3C_1 \times ^4C_1}{^7C_2} = \frac{3 \times 4}{21} = \frac{4}{7} \)
\( P\left(\frac{A}{E_1}\right) = \frac{6}{11}, P\left(\frac{A}{E_2}\right) = \frac{4}{11}, P\left(\frac{A}{E_3}\right) = \frac{5}{11} \)
\( \therefore P\left(\frac{E_2}{A}\right) = \frac{P(E_2) . P\left(\frac{A}{E_2}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right) + P(E_3) . P\left(\frac{A}{E_3}\right)} \)
\( = \frac{\frac{2}{7} \times \frac{4}{11}}{\frac{1}{7} \times \frac{6}{11} + \frac{2}{7} \times \frac{4}{11} + \frac{4}{7} \times \frac{5}{11}} = \frac{\frac{8}{77}}{\frac{6}{77} + \frac{8}{77} + \frac{20}{77}} = \frac{8}{77} \times \frac{77}{34} = \frac{4}{17} \)

Question. There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up head 75% of the times and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer: Let \( E_1, E_2, E_3 \) and \( A \) be event defined as:
\( E_1 = \text{selection of a two headed coin; } E_2 = \text{selection of a biased coin} \)
\( E_3 = \text{selection of an unbiased coin; } A = \text{coin shows head after tossing} \)
Now, \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
\( P\left(\frac{A}{E_1}\right) = 1, \quad P\left(\frac{A}{E_2}\right) = \frac{75}{100} = \frac{3}{4}, \quad P\left(\frac{A}{E_3}\right) = \frac{1}{2} \)
Here, required probability = \( P\left(\frac{E_1}{A}\right) \)
By using Baye’s theorem,
\( P\left(\frac{E_1}{A}\right) = \frac{P(E_1) . P\left(\frac{A}{E_1}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right) + P(E_3) . P\left(\frac{A}{E_3}\right)} = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}} \)
\( = \frac{\frac{1}{3}}{\frac{1}{3}\left(1 + \frac{3}{4} + \frac{1}{2}\right)} = \frac{1}{\frac{4 + 3 + 2}{4}} = \frac{4}{9} \)

Question. A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white? 
Answer: Let us define the following events.
\( E: \text{drawn balls are white}; \quad A: \text{2 white balls in bag.} \)
\( B: \text{3 white balls in bag}; \quad C: \text{4 white balls in bag.} \)
Then, \( P(A) = P(B) = P(C) = \frac{1}{3} \)
and \( P\left(\frac{E}{A}\right) = \frac{^2C_2}{^4C_2} = \frac{1}{6}, \quad P\left(\frac{E}{B}\right) = \frac{^3C_2}{^4C_2} = \frac{3}{6}, \quad P\left(\frac{E}{C}\right) = \frac{^4C_2}{^4C_2} = 1 \)
By applying Baye’s theorem
\( P\left(\frac{C}{E}\right) = \frac{P(C) . P\left(\frac{E}{C}\right)}{P(A) . P\left(\frac{E}{A}\right) + P(B) . P\left(\frac{E}{B}\right) + P(C) P\left(\frac{E}{C}\right)} \)
\( = \frac{\frac{1}{3} \times 1}{\left(\frac{1}{3} \times \frac{1}{6}\right) + \left(\frac{1}{3} \times \frac{3}{6}\right) + \left(\frac{1}{3} \times 1\right)} = \frac{1}{\frac{1}{6} + \frac{3}{6} + 1} = \frac{3}{5} \)

Question. An urn contains 4 white and 3 red balls. Let \( X \) be the number of red balls in a random draw of three balls. Find the mean and variance of \( X \). 
Answer: Let X be the number of red balls in a random draw of three balls.
As there are 3 red balls, possible values of X are 0, 1, 2, 3.
\( P(0) = \frac{^3C_0 \times ^4C_3}{^7C_3} = \frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{4}{35} \quad P(1) = \frac{^3C_1 \times ^4C_2}{^7C_3} = \frac{3 \times 6 \times 6}{7 \times 6 \times 5} = \frac{18}{35} \)
\( P(2) = \frac{^3C_2 \times ^4C_1}{^7C_3} = \frac{3 \times 4 \times 6}{7 \times 6 \times 5} = \frac{12}{35} \quad P(3) = \frac{^3C_3 \times ^4C_0}{^7C_3} = \frac{1 \times 1 \times 6}{7 \times 6 \times 5} = \frac{1}{35} \)
For calculation of Mean & Variance

Mean \( = \Sigma XP(X) = \frac{9}{7} \)
Variance \( = \Sigma X^2 . P(X) - (\Sigma X . P(X))^2 = \frac{15}{7} - \frac{81}{49} = \frac{24}{49} \)

Question. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75 metres. Find the probability that the selected student is a girl. 
Answer: Let \( E_1, E_2, A \) be events such that
\( E_1 = \text{student selected is girl}; \quad E_2 = \text{student selected is Boy} \)
\( A = \text{student selected is taller than 1.75 metres.} \)
Here \( P\left(\frac{E_1}{A}\right) \) is required.
Now, \( P(E_1) = \frac{60}{100} = \frac{3}{5}, \quad P(E_2) = \frac{40}{100} = \frac{2}{5} \quad P\left(\frac{A}{E_1}\right) = \frac{1}{100}, \quad P\left(\frac{A}{E_2}\right) = \frac{4}{100} \)
\( P\left(\frac{E_1}{A}\right) = \frac{P(E_1) . P\left(\frac{A}{E_1}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right)} \)
\( = \frac{\frac{3}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{1}{100} + \frac{2}{5} \times \frac{4}{100}} = \frac{\frac{3}{500}}{\frac{3}{500} + \frac{8}{500}} = \frac{3}{500} \times \frac{500}{11} = \frac{3}{11} \)

Question. A factory has two machines \( A \) and \( B \). Past record shows that machine \( A \) produced 60% of the items of output and machine \( B \) produced 40% of the items. Further, 2% of the items produced by machine \( A \) and 1% produced by machine \( B \) were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine \( B \)?
Answer: Let \( E_1, E_2 \) and \( A \) be event such that
\( E_1 = \text{Production of items by machine } A \)
\( E_2 = \text{Production of items by machine } B \)
\( A = \text{Selection of defective items.} \)
\( P(E_1) = \frac{60}{100} = \frac{3}{5}, P(E_2) = \frac{40}{100} = \frac{2}{5}, \quad P\left(\frac{A}{E_1}\right) = \frac{2}{100} = \frac{1}{50}, P\left(\frac{A}{E_2}\right) = \frac{1}{100} \)
\( P\left(\frac{E_2}{A}\right) \) is required
By Baye’s theorem
\( P\left(\frac{E_2}{A}\right) = \frac{P(E_2) . P\left(\frac{A}{E_2}\right)}{P(E_1) . P\left(\frac{A}{E_1}\right) + P(E_2) . P\left(\frac{A}{E_2}\right)} \)
\( = \frac{\frac{2}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{1}{50} + \frac{2}{5} \times \frac{1}{100}} = \frac{\frac{2}{500}}{\frac{3}{250} + \frac{2}{500}} = \frac{2}{500} \times \frac{500}{6 + 2} = \frac{1}{4} \)

Question. Two groups are competing for the position on the Board of Directors of a corporation. The probabilities that the first and the second group will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3, if the second group wins. Find the probability that the new product was introduced by the second group. 
Answer: \( P(G_I) = 0.6, \quad P(G_{II}) = 0.4 \)
Let E is the event of introducing new product then
\( P(E/G_I) = 0.7 \quad P(E/G_{II}) = 0.3 \)
To find \( P(G_{II}/E) \)
Using Baye’s theorem, we get
\( P(G_{II}/E) = \frac{P(G_{II}) . P(E/G_{II})}{P(G_I) . P(E/G_I) + P(G_{II}) . P(E/G_{II})} \)
\( = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{12}{54} = \frac{2}{9} \)