CBSE Class 12 Mathematics HOTs Probability Set 02

Very Short Answer Type Questions

Question. Four distinct numbers are chosen randomly from the first 60 natural numbers. Find the probability that all the four numbers are divisible by both 3 and 4.
Answer: Total numbers from 1 to 60 which are divisible by both 3 and 4 (i.e. divisible by 12) is \( \{12, 24, 36, 48, 60\} \).
Number of such numbers = 5
Total numbers = 60
Number of ways of choosing 4 numbers from 5 divisible by 12 is \( {}^{5}\text{C}_{4} \).
Total number of ways of choosing 4 numbers from 60 is \( {}^{60}\text{C}_{4} \).
Required probability = \( \frac{{}^{5}\text{C}_{4}}{{}^{60}\text{C}_{4}} \)

Question. There are 40 tickets bearing numbers from 1 to 40. One ticket is drawn at random. Find the probability that the number on it is a multiple of 3 or 5.
Answer: Let \( E \) = Event of getting a multiple of 3.
\( F \) = Event of getting a multiple of 5.
and \( E \cap F \) = Event of getting a multiple of 3 and 5 i.e. 15.
\( \therefore P(E) = \frac{13}{40} \) [\( \because E = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39\} \)]
\( P(F) = \frac{8}{40} \) [\( \because F = \{5, 10, 15, 20, 25, 30, 35, 40\} \)]
and \( P(E \cap F) = \frac{2}{40} \) [\( \because E \cap F = \{15, 30\} \)]
Required probability,
\( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)
\( = \frac{13}{40} + \frac{8}{40} - \frac{2}{40} \)
\( = \frac{13 + 8 - 2}{40} = \frac{19}{40} \)

Question. The probability that A hits the target is \( \frac{1}{3} \) and the probability that B hits it is \( \frac{2}{5} \). If both try to hit the target independently, find the probability that the target is hit.
Answer: Let \( E \) and \( F \) be the events defined as follows:
\( E \): A hits the target
and \( F \): B hits the target
\( P(E) = \frac{1}{3} \), \( P(\overline{E}) = 1 - P(E) = \frac{2}{3} \)
\( P(F) = \frac{2}{5} \), \( P(\overline{F}) = 1 - P(F) = \frac{3}{5} \)
\( P(\text{the target is hit}) = 1 - P(\text{the target is not hit}) \)
\( = 1 - (P(\overline{E}) \times P(\overline{F})) \)
\( = 1 - \left(\frac{2}{3} \times \frac{3}{5}\right) = 1 - \frac{6}{15} \)
\( = \frac{9}{15} = \frac{3}{5} \)

Question. The probabilities of A and B solving a problem independently are \( \frac{1}{3} \) and \( \frac{1}{4} \), respectively. If both of them try to solve the problem independently, then what is the probability that the problem is solved?
Answer: Let \( A \) and \( B \) be the respective events of solving the problem by A and B, respectively.
Then, \( P(A) = \frac{1}{3} \) and \( P(B) = \frac{1}{4} \)
\( \therefore \) Required probability \( = 1 - P(\overline{A})P(\overline{B}) \)
\( = 1 - \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \)
\( = 1 - \frac{2}{3} \times \frac{3}{4} \)
\( = 1 - \frac{1}{2} = \frac{1}{2} \)

Question. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week.
Answer: Let \( E \) be the event that it will rain on any particular day.
\( \therefore P(E) = 50\% = \frac{1}{2} \Rightarrow P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \therefore \) Required probability
\( = P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(\overline{E}) \cdot P(\overline{E}) \cdot P(\overline{E}) \)
\( = (P(E))^4 (P(\overline{E}))^3 \)
\( = \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^3 = \frac{1}{128} \)

Question. If A and B are two events such that \( P(A) = 0.4 \), \( P(B) = 0.8 \) and \( P(B/A) = 0.6 \), then find \( P(A/B) \).
Answer: We have, \( P(A) = 0.4 \), \( P(B) = 0.8 \) and \( P(B/A) = 0.6 \)
We know that
\( P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)} \)
\( \Rightarrow 0.6 = \frac{P(A \cap B)}{0.4} \)
\( \therefore P(A \cap B) = 0.6 \times 0.4 = 0.24 \)
Now, \( P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.8} = 0.3 \)

Question. Evaluate \( P(A \cup B) \), if \( 2P(A) = P(B) = \frac{5}{13} \) and \( P(A/B) = \frac{2}{5} \).
Answer: We have, \( 2P(A) = P(B) = \frac{5}{13} \)
\( \Rightarrow P(A) = \frac{5}{26} \), \( P(B) = \frac{5}{13} \)
and \( P(A/B) = \frac{2}{5} \)
\( \because P\left(\frac{A}{B}\right) = \frac{P(A \cap B)}{P(B)} \)
\( \Rightarrow \frac{2}{5} = \frac{P(A \cap B)}{5/13} \)
\( \Rightarrow P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13} \)
\( \because P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \)
\( = \frac{5 + 10 - 4}{26} = \frac{11}{26} \)

Question. Given that E and F are events such that \( P(E) = 0.6 \), \( P(F) = 0.3 \) and \( P(E \cap F) = 0.2 \). Find \( P(E/F) \) and \( P(F/E) \).
Answer: We know that \( P\left(\frac{E}{F}\right) = \frac{P(E \cap F)}{P(F)} \) and \( P\left(\frac{F}{E}\right) = \frac{P(E \cap F)}{P(E)} \)
Now, \( P\left(\frac{E}{F}\right) = \frac{0.2}{0.3} = \frac{2}{3} \)
and \( P\left(\frac{F}{E}\right) = \frac{0.2}{0.6} = \frac{1}{3} \)

Question. A speaks truth in 80% cases and B speaks truth in 90% cases. In what percentage of cases are they likely to agree with each other in stating the same fact?
Answer: Let \( A_T \): Event that A speaks truth and \( B_T \): Event that B speaks truth.
Given, \( P(A_T) = \frac{80}{100} = \frac{4}{5} \), \( P(B_T) = \frac{90}{100} = \frac{9}{10} \)
\( P(\text{agree}) = P(\text{both speaking truth or both telling lie}) \)
\( = P(A_T \cap B_T) \text{ or } P(\overline{A}_T \cap \overline{B}_T) \)
\( = P(A_T)P(B_T) + P(\overline{A}_T)P(\overline{B}_T) \)
\( = \left(\frac{4}{5}\right)\left(\frac{9}{10}\right) + \left(\frac{1}{5}\right)\left(\frac{1}{10}\right) \)
\( = \frac{36 + 1}{50} = \frac{37}{50} \)
\( = \frac{74}{100} = 74\% \)

Question. If \( P(A) = \frac{4}{5} \) and \( P(A \cap B) = \frac{7}{10} \), then find \( P(B/A) \).
Answer: Given, \( P(A) = \frac{4}{5} \) and \( P(A \cap B) = \frac{7}{10} \)
We know that \( P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)} \)
\( \Rightarrow P\left(\frac{B}{A}\right) = \frac{7/10}{4/5} = \frac{7}{10} \times \frac{5}{4} = \frac{7}{8} \)

Short Answer Type Questions

Question. If A and B are two events such that \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{6} \) and \( P(A \cap B) = \frac{1}{7} \), then find (i) \( P(A \cup B) \) (ii) \( P(\overline{B}/\overline{A}) \).
Answer: Given, \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{6} \) and \( P(A \cap B) = \frac{1}{7} \)
(i) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{1}{4} + \frac{1}{6} - \frac{1}{7} \)
\( = \frac{21 + 14 - 12}{84} = \frac{23}{84} \)
(ii) \( P\left(\frac{\overline{B}}{\overline{A}}\right) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})} \)
\( = \frac{P(\overline{A \cup B})}{1 - P(A)} = \frac{1 - P(A \cup B)}{1 - P(A)} \)
\( = \frac{1 - \frac{23}{84}}{1 - \frac{1}{4}} = \frac{\frac{61}{84}}{\frac{3}{4}} \)
\( = \frac{61}{84} \times \frac{4}{3} = \frac{61}{63} \)

Question. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2 and 0.1, respectively. What is the probability that atleast one shot hits the plane?
Answer: Let \( E_1, E_2, E_3, E_4 \) be the respective events that the plane is hit in the 1st, 2nd, 3rd and 4th shot. Then,
\( P(E_1) = 0.4, P(E_2) = 0.3, P(E_3) = 0.2 \) and \( P(E_4) = 0.1 \)
\( \therefore P(\overline{E}_1) = (1 - 0.4) = 0.6, P(\overline{E}_2) = (1 - 0.3) = 0.7 \),
\( P(\overline{E}_3) = (1 - 0.2) = 0.8, P(\overline{E}_4) = (1 - 0.1) = 0.9 \).
\( \therefore P(\text{atleast one shot hits the plane}) \)
\( = 1 - P(\overline{E}_1 \text{ and } \overline{E}_2 \text{ and } \overline{E}_3 \text{ and } \overline{E}_4) \)
\( = 1 - \{P(\overline{E}_1) \times P(\overline{E}_2) \times P(\overline{E}_3) \times P(\overline{E}_4)\} \)
\( = 1 - \{0.6 \times 0.7 \times 0.8 \times 0.9\} = 1 - \{0.3024\} = 0.6976 \)
Hence, the probability that atleast one shot hits the plane is 0.6976.

Question. A speaks truth in 75% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? Do you think that statement of B is true?
Answer: Let \( A_T \): Event that A speaks truth and \( B_T \): Event that B speaks truth.
Given, \( P(A_T) = \frac{75}{100} \), then \( P(\overline{A}_T) = 1 - \frac{75}{100} = \frac{25}{100} \)
and \( P(B_T) = \frac{90}{100} \), then \( P(\overline{B}_T) = 1 - \frac{90}{100} = \frac{10}{100} \).
Now, \( P(\text{A and B are contradict to each other}) \)
\( = P(A_T \cap \overline{B}_T) + P(\overline{A}_T \cap B_T) \)
\( = P(A_T) \cdot P(\overline{B}_T) + P(\overline{A}_T) \cdot P(B_T) \) [\( \because A_T \) and \( B_T \) are independent events]
\( = \frac{75}{100} \times \frac{10}{100} + \frac{25}{100} \times \frac{90}{100} \)
\( = \frac{750 + 2250}{10000} = \frac{3000}{10000} = \frac{3}{10} \)
\( \therefore \) Percentage of \( P(\text{A and B are contradict to each other}) = \frac{3}{10} \times 100 = 30\% \).
Since, B speaks truth in only 90% (i.e. not 100%) of the cases, therefore we think, the statement of B may be false.

Question. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact? Do you think, when they agree, means both are speaking truth?
Answer: Let \( P_T \): Event that P speaks truth and \( Q_T \): Event that Q speaks truth.
Given, \( P(P_T) = \frac{70}{100} \), then \( P(\overline{P}_T) = 1 - \frac{70}{100} = \frac{30}{100} \)
and \( P(Q_T) = \frac{80}{100} \), then \( P(\overline{Q}_T) = 1 - \frac{80}{100} = \frac{20}{100} \).
\( P(\text{A and B are agree to each other}) \)
\( = P(P_T \cap Q_T) + P(\overline{P}_T \cap \overline{Q}_T) \)
\( = P(P_T) \cdot P(Q_T) + P(\overline{P}_T) \cdot P(\overline{Q}_T) \) [\( \because P_T \) and \( Q_T \) are independent events]
\( = \frac{70}{100} \times \frac{80}{100} + \frac{30}{100} \times \frac{20}{100} \)
\( = \frac{5600 + 600}{10000} = \frac{6200}{10000} = \frac{62}{100} \)
\( \therefore \) Percentage of \( P(\text{A and B are agree to each other}) = \frac{62}{100} \times 100 = 62\% \).
No, agree does not mean that they are speaking truth.

Question. A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?
Answer: Let \( A \) and \( B \) denote the events A speaks the truth and B speaks the truth, respectively.
We have, \( P(A) = \frac{60}{100} \Rightarrow P(\overline{A}) = 1 - P(A) = 1 - \frac{60}{100} = \frac{40}{100} \)
and \( P(B) = \frac{90}{100} \Rightarrow P(\overline{B}) = 1 - P(B) = 1 - \frac{90}{100} = \frac{10}{100} \)
So, required probability \( = P(A)P(\overline{B}) + P(\overline{A})P(B) \)
\( = \frac{60}{100} \times \frac{10}{100} + \frac{40}{100} \times \frac{90}{100} \)
\( = \frac{600 + 3600}{10000} = \frac{4200}{10000} = 42\% \)
Hence, they are likely to contradict each other in 42% of cases in stating the same fact.
Since, no one trusts a liar, so statement of B will carry more weight as he speaks the truth in more number of cases than A.

Question. An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw' and B be the event ‘odd number on the second throw’. Check the independence of the events A and B.
Answer: On throwing a die twice, we get 36 elementary events of the experiment which are to be considered to be equally likely. Odd numbers are 1, 3 and 5.
\( P(A) = P(\text{odd number on the first throw}) = \frac{18}{36} = \frac{1}{2} \)
[\( \because \) possible outcomes are \( (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5) \text{ and } (5, 6) \) i.e. 18 outcomes]
Similarly, \( P(B) = P(\text{odd number on the second throw}) = \frac{18}{36} = \frac{1}{2} \)
Also, \( P(A \cap B) = P(\text{odd number on both throws}) = \frac{9}{36} = \frac{1}{4} \)
[\( \because \) events are \( (1,1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3) \text{ and } (5, 5) \) i.e. 9 outcomes]
Now, \( P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
Thus, \( P(A \cap B) = P(A) \cdot P(B) \)
Hence, A and B are independent events.

Question. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘number is odd’ and B be the event, ‘number is green’. Are A and B independent?
Answer: When a die is thrown, then sample space is
\( S = \{1, 2, 3, 4, 5, 6\} \Rightarrow n(S) = 6 \)
Also, A: number is odd and B: number is green.
\( A = \{1, 3, 5\} \) and \( B = \{4, 5, 6\} \)
So, \( A \cap B = \{5\} \)
\( \Rightarrow n(A) = 3, n(B) = 3 \) and \( n(A \cap B) = 1 \)
Now, \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}, P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2} \)
and \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{6} \)
Now, \( P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \neq P(A \cap B) \)
\( \therefore P(A \cap B) \neq P(A) \times P(B) \)
Thus, A and B are not independent events.

Question. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, then what is the conditional probability that both are girls? Given that (i) the youngest is a girl? (ii) atleast one is a girl?
Answer: Let \( B \) and \( b \) represent elder and younger boy child, respectively. Also, \( G \) and \( g \) represent elder and younger girl child, respectively. If a family has two children, then all possible cases are
\( S = \{Bb, Bg, Gg, Gb\} \)
\( n(S) = 4 \)
Let us define event A: Both children are girls, then
\( A = \{Gg\} \Rightarrow n(A) = 1 \)
(i) Let \( E_1 \): The event that youngest child is a girl.
Then, \( E_1 = \{Bg, Gg\} \Rightarrow n(E_1) = 2 \)
So, \( P(E_1) = \frac{n(E_1)}{n(S)} = \frac{2}{4} = \frac{1}{2} \)
and \( A \cap E_1 = \{Gg\} \Rightarrow n(A \cap E_1) = 1 \)
So, \( P(A \cap E_1) = \frac{n(A \cap E_1)}{n(S)} = \frac{1}{4} \)
Now, \( P\left(\frac{A}{E_1}\right) = \frac{P(A \cap E_1)}{P(E_1)} = \frac{1/4}{1/2} = \frac{1}{2} \)
\( \therefore \) Required probability \( = \frac{1}{2} \)
(ii) Let \( E_2 \): The event that atleast one is girl.
Then, \( E_2 = \{Bg, Gg, Gb\} \Rightarrow n(E_2) = 3 \)
So, \( P(E_2) = \frac{n(E_2)}{n(S)} = \frac{3}{4} \)
and \( (A \cap E_2) = \{Gg\} \Rightarrow n(A \cap E_2) = 1 \)
So, \( P(A \cap E_2) = \frac{n(A \cap E_2)}{n(S)} = \frac{1}{4} \)
Now, \( P\left(\frac{A}{E_2}\right) = \frac{P(A \cap E_2)}{P(E_2)} = \frac{1/4}{3/4} = \frac{1}{3} \)
\( \therefore \) Required probability \( = \frac{1}{3} \)

Question. Let 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, then find the probability of one of them being red and another black.
Answer: Given, bag A = 4 black and 6 red balls
and bag B = 7 black and 3 red balls.
Let \( E_1 \) = The event that die show 1 or 2
\( E_2 \) = The event that die show 3, 4, 5 or 6
and \( E \) = The event that among two drawn balls, one of them is red and other is black
Now, \( P(E_1) = \frac{2}{6}, P(E_2) = \frac{4}{6} \) [\( \because \) total number in a die is six]
\( P\left(\frac{E}{E_1}\right) = P(\text{getting one red and one black from bag A}) \)
\( = \frac{{}^{4}\text{C}_{1} \times {}^{6}\text{C}_{1}}{{}^{10}\text{C}_{2}} = \frac{4 \times 6 \times 2}{10 \times 9} \)
and \( P\left(\frac{E}{E_2}\right) = P(\text{getting one red and one black from bag B}) \)
\( = \frac{{}^{7}\text{C}_{1} \times {}^{3}\text{C}_{1}}{{}^{10}\text{C}_{2}} = \frac{7 \times 3 \times 2}{10 \times 9} \)
\( \therefore \) Total probability,
\( P(E) = P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right) \)
\( = \frac{2}{6}\left(\frac{4 \times 6 \times 2}{10 \times 9}\right) + \frac{4}{6}\left(\frac{7 \times 3 \times 2}{10 \times 9}\right) \)
\( = \frac{4 \times 6}{6 \times 10 \times 9}(4 + 7) = \frac{4 \times 6 \times 11}{6 \times 10 \times 9} = \frac{22}{45} \)

Question. A bag contains (2n + 1) coins. It is known that n of these coins have a head on both its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is \( \frac{31}{42} \), then find the value of n.
Answer: Given, total number of coins \( = 2n + 1 \)
Number of coins having head on both sides \( = n \)
Number of fair coins \( = (2n + 1) - n = n + 1 \)
Let \( E_1 \): Event of a coin having head on both sides is selected,
\( E_2 \): Event of a fair coin is selected
and \( A \): Event of head comes up in tossing a selected coin.
Then, \( P(E_1) = \frac{n}{2n + 1}; P(E_2) = \frac{n + 1}{2n + 1} \);
\( P(A/E_1) = 1; P(A/E_2) = \frac{1}{2} \)
It is given that \( P(A) = \frac{31}{42} \).
So, \( P(E_1)P(A/E_1) + P(E_2)P(A/E_2) = \frac{31}{42} \)
\( \Rightarrow \frac{n}{2n + 1} \times 1 + \frac{n + 1}{2n + 1} \times \frac{1}{2} = \frac{31}{42} \)
\( \Rightarrow \frac{1}{2n + 1} \left[n + \frac{n+1}{2}\right] = \frac{31}{42} \)
\( \Rightarrow 42(3n + 1) = 62(2n + 1) \)
\( \Rightarrow 126n + 42 = 124n + 62 \)
\( \Rightarrow 2n = 20 \)
\( \Rightarrow n = 10 \)

Question. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer: Let \(E_1\) = Red ball is transferred from Bag I to Bag II and \(E_2\) = Black ball is transferred from Bag I to Bag II.
\(\therefore P(E_1) = \frac{3}{3+4} = \frac{3}{7}\) and \(P(E_2) = \frac{4}{3+4} = \frac{4}{7}\).
\(P\left(\frac{E}{E_1}\right) = \frac{4+1}{(4+1)+5} = \frac{5}{10} = \frac{1}{2}\) and \(P\left(\frac{E}{E_2}\right) = \frac{4}{4+(5+1)} = \frac{4}{10} = \frac{2}{5}\).
\(\therefore P\left(\frac{E_2}{E}\right) = \frac{P\left(\frac{E}{E_2}\right) \cdot P(E_2)}{P\left(\frac{E}{E_1}\right) \cdot P(E_1) + P\left(\frac{E}{E_2}\right) \cdot P(E_2)}\)
\(= \frac{\frac{2}{5} \times \frac{4}{7}}{\frac{1}{2} \times \frac{3}{7} + \frac{2}{5} \times \frac{4}{7}} = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{16}{31}\).
Ans. \(\frac{16}{31}\)

Question. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die?
Answer: Let \(E_1\) be the event that the girl gets 1 or 2, \(E_2\) be the event that the girl gets 3, 4, 5 or 6, and \(A\) be the event that the girl gets exactly a ‘tail’.
Then, \(P(E_1) = \frac{2}{6} = \frac{1}{3}\) and \(P(E_2) = \frac{4}{6} = \frac{2}{3}\).
\(P\left(\frac{A}{E_1}\right)\) = P (getting exactly one tail when a coin is tossed three times) = \(\frac{3}{8}\).
\(P\left(\frac{A}{E_2}\right)\) = P (getting exactly a tail when a coin is tossed once) = \(\frac{1}{2}\).
Now, required probability:
\(P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)}\)
\(= \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{8}{11}\).

Question. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, then what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer: Let \(E_1\) = Event that 5 or 6 is shown on die, and \(E_2\) = Event that 1, 2, 3 or 4 is shown on die.
Here, \(n(E_1) = 2\), \(n(E_2) = 4\) and \(n(S) = 6\).
\(\therefore P(E_1) = \frac{2}{6} = \frac{1}{3}\) and \(P(E_2) = \frac{4}{6} = \frac{2}{3}\).
Let \(E\) = The event that exactly one head shows up.
\(P\left(\frac{E}{E_1}\right)\) = P (exactly one head shows up when coin is tossed thrice) = \(P[HTT, THT, TTH] = \frac{3}{8}\) (since total number of outcomes = \(2^3 = 8\)).
\(P\left(\frac{E}{E_2}\right)\) = P (head shows up when coin is tossed once) = \(\frac{1}{2}\).
The probability that the girl threw 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by:
\(P\left(\frac{E_2}{E}\right) = \frac{P(E_2) \cdot P\left(\frac{E}{E_2}\right)}{P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right)}\)
\(= \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{8}{11}\).
Ans. \(\frac{8}{11}\)

Question. Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and 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 introduced was by the second group.
Answer: Let \(E_1\) and \(E_2\) denote the events that the first and second group will win. Then, \(P(E_1) = 0.6\) and \(P(E_2) = 0.4\).
Let \(E\) be the event of introducing the new product.
Then, \(P\left(\frac{E}{E_1}\right) = 0.7\) and \(P\left(\frac{E}{E_2}\right) = 0.3\).
Now, we have to find the probability that the new product is introduced by the second event:
\(\therefore P\left(\frac{E_2}{E}\right) = \frac{P(E_2) \cdot P\left(\frac{E}{E_2}\right)}{P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right)}\)
\(= \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = 0.22\).

Question. Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III?
Answer: Let \(E_1, E_2, E_3\) and \(A\) denote the following events:
\(E_1\) = Bag I is chosen, \(E_2\) = Bag II is chosen, \(E_3\) = Bag III is chosen,
\(A\) = The balls drawn from the chosen bag are white and red.
Since, one of the bags is chosen at random:
\(\therefore P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}\).
If one white and red balls are chosen from Bag I with replacement:
Then, \(P\left(\frac{A}{E_1}\right) = \frac{1}{6} \times \frac{3}{6} \times 2\).
Similarly, \(P\left(\frac{A}{E_2}\right) = \frac{2}{4} \times \frac{1}{4} \times 2\) and \(P\left(\frac{A}{E_3}\right) = \frac{4}{9} \times \frac{2}{9} \times 2\).
By Bayes’ theorem,
Required probability = \(P\left(\frac{E_3}{A}\right) = \frac{P(E_3) \times P\left(\frac{A}{E_3}\right)}{\sum_{i=1}^{3} P(E_i) \times P\left(\frac{A}{E_i}\right)}\)
\(= \frac{\frac{1}{3} \times \frac{4}{9} \times \frac{2}{9} \times 2}{\frac{1}{3} \times \frac{1}{6} \times \frac{3}{6} \times 2 + \frac{1}{3} \times \frac{2}{4} \times \frac{1}{4} \times 2 + \frac{1}{3} \times \frac{4}{9} \times \frac{2}{9} \times 2} = \frac{64}{199}\).

Question. A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.
Answer: Let us define the following events:
\(E_1\): Bag X is selected
\(E_2\): Bag Y is selected
and \(E\): Getting one white and one black ball in a draw of two balls.
Here, \(P(E_1) = P(E_2) = \frac{1}{2}\) [since probability of selecting each bag is equal].
Now, \(P\left(\frac{E}{E_1}\right) = \text{Probability of drawing one white and one black ball from Bag X} = \frac{^{4}C_1 \times ^{2}C_1}{^{6}C_2} = \frac{4 \times 2}{\frac{6 \times 5}{2 \times 1}} = \frac{8}{15}\).
And \(P\left(\frac{E}{E_2}\right) = \text{Probability of drawing one white and one black ball from Bag Y} = \frac{^{3}C_1 \times ^{3}C_1}{^{6}C_2} = \frac{3 \times 3}{15} = \frac{3}{5}\).
\(\therefore\) The probability that the one white and one black balls are drawn from Bag Y:
\(P\left(\frac{E_2}{E}\right) = \frac{P(E_2) \cdot P\left(\frac{E}{E_2}\right)}{P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right)}\) [using Bayes’ theorem]
\(= \frac{\frac{1}{2} \times \frac{3}{5}}{\frac{1}{2} \times \frac{8}{15} + \frac{1}{2} \times \frac{3}{5}} = \frac{\frac{3}{5}}{\frac{8}{15} + \frac{3}{5}} = \frac{9}{17}\).

Question. Three machines \(E_1, E_2\) and \(E_3\) in a certain factory producing electric bulbs, produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines \(E_1\) and \(E_2\) are defective and that 5% of those produced by machine \(E_3\) are defective. If one bulb is picked up at random from a day's production, calculate the probability that it is defective.
Answer: Let \(A_1\): Event that the bulb is produced by machine \(E_1\),
\(A_2\): Event that the bulb is produced by machine \(E_2\),
\(A_3\): Event that the bulb is produced by machine \(E_3\),
\(A\): Event that the picked up bulb is defective.
Here, \(P(A_1) = 50\% = \frac{1}{2}\), \(P(A_2) = 25\% = \frac{1}{4}\) and \(P(A_3) = 25\% = \frac{1}{4}\).
Also, \(P\left(\frac{A}{A_1}\right) = 4\% = \frac{1}{25}\), \(P\left(\frac{A}{A_2}\right) = 4\% = \frac{1}{25}\) and \(P\left(\frac{A}{A_3}\right) = 5\% = \frac{1}{20}\).
\(\dots\) The probability that the picked bulb is defective:
\(P(A) = P(A_1) \times P\left(\frac{A}{A_1}\right) + P(A_2) \times P\left(\frac{A}{A_2}\right) + P(A_3) \times P\left(\frac{A}{A_3}\right)\)
\(= \frac{1}{2} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{20}\)
\(= \frac{1}{50} + \frac{1}{100} + \frac{1}{80} = \frac{8 + 4 + 5}{400} = \frac{17}{400} = 0.0425\).

Long Answer Type Questions

Question. The odds against a man who is 45 yr old, living till he is 70 are 7 : 5 and the odds against his wife who is now 36, living till she is 61 are 5 : 3. Find the probability that (i) the couple will be alive 25 yr hence, (ii) atleast one of them will be alive 25 yr hence. 
Answer: Let \(E_1\) = Event that the husband will be alive 25 yr hence, and \(E_2\) = Event that the wife will be alive 25 yr hence.
Then, \(P(E_1) = \frac{5}{12}\) and \(P(E_2) = \frac{3}{8}\).
\(\therefore P(\overline{E_1}) = \left(1 - \frac{5}{12}\right) = \frac{7}{12}\) and \(P(\overline{E_2}) = \left(1 - \frac{3}{8}\right) = \frac{5}{8}\).
Clearly, \(E_1\) and \(E_2\) are independent events.
(i) P(the couple will be alive 25 yr hence)
\(= P(E_1 \text{ and } E_2) = P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) = \frac{5}{12} \times \frac{3}{8} = \frac{5}{32}\).
(ii) P(atleast one of them will be alive 25 yr hence)
\(= 1 - P(\text{none will be alive 25 yr hence})\)
\(= 1 - P[(\text{not } E_1) \text{ and } (\text{not } E_2)] = 1 - P[\overline{E_1} \cap \overline{E_2}]\)
\(= 1 - P(\overline{E_1}) \cdot P(\overline{E_2})\) [\(\because \overline{E_1}\) and \(\overline{E_2}\) are independent events]
\(= 1 - \left(\frac{7}{12} \times \frac{5}{8}\right) = 1 - \frac{35}{96} = \frac{61}{96}\).

Question. A manufacturer has three machine operators A, B and C. The first operator A produces 1% of defective items, whereas the other two operators B and C produces 5% and 7% defective items, respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job 20% of the time. All the items are put into one stockpile and then one items is chosen at random from this and is found to be defective. What is the probability that it was produced by A?
Answer: Let \(A\): Event that item produced by Operator A,
\(B\): Event that item produced by Operator B,
\(C\): Event that item produced by Operator C,
\(D\): Event that item produced is defective.
We need to find out the probability that item is produced by Operator A if it is defective i.e., \(P(A / D)\).
So, \(P(A / D) = \frac{P(A) \cdot P(D / A)}{P(A) \cdot P(D / A) + P(B) \cdot P(D / B) + P(C) \cdot P(D / C)}\) [by Bayes’ theorem]
\(P(A)\) = Probability of item produced by Operator A = \(50\% = 0.5\).
\(P(B)\) = Probability of item produced by Operator B = \(30\% = 0.3\).
\(P(C)\) = Probability of item produced by Operator C = \(20\% = 0.2\).
\(P(D / A)\) = Probability of a defective item produced by Operator A = \(1\% = 0.01\).
\(P(D / B)\) = Probability of a defective item produced by Operator B = \(5\% = 0.05\).
\(P(D / C)\) = Probability of a defective item produced by Operator C = \(7\% = 0.07\).
On putting these values, we get:
\(P(A / D) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07}\)
\(= \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34}\).
Therefore, required probability is \(\frac{5}{34}\).

Question. A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.
Answer: Let \(E_1\) = Event that 1 occurs in a die,
\(E_2\) = Event that 1 does not occur in a die,
\(A\) = Event that the man reports that 1 occurs in a die.
Then, \(P(E_1) = \frac{1}{6}\) and \(P(E_2) = \frac{5}{6}\).
\(P\text{(man reports that 1 occurs when 1 occur)} = P\left(\frac{A}{E_1}\right) = \frac{3}{5}\),
and \(P\text{(man reports that 1 occurs but 1 does not occur)} = P\left(\frac{A}{E_2}\right) = \frac{2}{5}\).
Thus, by Bayes’ theorem, we get \(P\text{(get actually 1 when he reports that 1 occur)}\):
\(P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)}\)
\(= \frac{\frac{1}{6} \times \frac{3}{5}}{\frac{1}{6} \times \frac{3}{5} + \frac{5}{6} \times \frac{2}{5}} = \frac{\frac{3}{30}}{\frac{3}{30} + \frac{10}{30}} = \frac{3}{13}\).

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 us define the following events:
\(E_1\) = Lost card is a spade card,
\(E_2\) = Lost card is not a spade card,
and \(A\) = Drawn cards are spade cards.
Then, \(P(E_1) = \frac{13}{52} = \frac{1}{4}\), \(P(E_2) = \frac{39}{52} = \frac{3}{4}\).
\(P(A / E_1) = \frac{^{12}C_3}{^{51}C_3} = \frac{220}{20825}\), and \(P(A / E_2) = \frac{^{13}C_3}{^{51}C_3} = \frac{286}{20825}\).
Now, required probability, \(P(E_1 / A)\):
\(P(E_1 / A) = \frac{P(E_1) \cdot P(A / E_1)}{P(E_1) \cdot P(A / E_1) + P(E_2) \cdot P(A / E_2)}\)
\(= \frac{\frac{1}{4} \times \frac{220}{20825}}{\frac{1}{4} \times \frac{220}{20825} + \frac{3}{4} \times \frac{286}{20825}} = \frac{220}{220 + 858} = \frac{220}{1078} = \frac{20}{98} = \frac{10}{49}\).

Question. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers, the probability of their meeting an accident respectively are 0.01, 0.03 and 0.15. One of the insured persons meets with an accident. What is the probability that he is a car driver? 
Answer: Given, number of scooter drivers = 2000,
Number of car drivers = 4000,
Number of truck drivers = 6000.
Let \(A\): Insured person is a scooter driver,
\(B\): Insured person is a car driver,
\(C\): Insured person is a truck driver,
\(D\): The insured person meets with an accident.
\(\therefore P(A) = \frac{2000}{2000 + 4000 + 6000} = \frac{1}{6}\).
\(P(B) = \frac{4000}{2000 + 4000 + 6000} = \frac{1}{3}\).
\(P(C) = \frac{6000}{2000 + 4000 + 6000} = \frac{1}{2}\).
Also, \(P\left(\frac{D}{A}\right) = 0.01\), \(P\left(\frac{D}{B}\right) = 0.03\), and \(P\left(\frac{D}{C}\right) = 0.15\).
By Bayes’ theorem,
\(P\left(\frac{B}{D}\right) = \frac{P(B) \times P\left(\frac{D}{B}\right)}{P(A) \times P\left(\frac{D}{A}\right) + P(B) \times P\left(\frac{D}{B}\right) + P(C) \times P\left(\frac{D}{C}\right)}\)
\(= \frac{\frac{1}{3} \times 0.03}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15}\)
\(= \frac{\frac{3}{300}}{\frac{1}{600} + \frac{3}{300} + \frac{15}{200}} = \frac{3}{300} \times \frac{600}{52} = \frac{6}{52} = \frac{3}{26}\).
Hence, the probability that he is a car driver is \(\frac{3}{26}\).

Question. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed and it shows head. What is the probability that it was the two headed coin?
Answer: Given that there are 3 coins.
Let these are \(A, B\) and \(C\), respectively.
\(\therefore P(A) = P(B) = P(C) = \frac{1}{3}\).
Let \(H\) be the event that the coin shows heads.
For coin \(A\), probability of getting head = \(P\left(\frac{H}{A}\right) = 1\).
For coin \(B\), probability of getting head = \(P\left(\frac{H}{B}\right) = \frac{3}{4}\).
For coin \(C\), probability of getting head = \(P\left(\frac{H}{C}\right) = 0.6 = \frac{3}{5}\).
We have to find \(P\left(\frac{A}{H}\right)\) = Probability of getting \(H\) by coin \(A\).
By Bayes’ theorem,
\(P\left(\frac{A}{H}\right) = \frac{P\left(\frac{H}{A}\right) \cdot P(A)}{P\left(\frac{H}{A}\right) \cdot P(A) + P\left(\frac{H}{B}\right) \cdot P(B) + P\left(\frac{H}{C}\right) \cdot P(C)}\)
\(= \frac{1 \times \frac{1}{3}}{1 \times \frac{1}{3} + \frac{3}{4} \times \frac{1}{3} + \frac{3}{5} \times \frac{1}{3}} = \frac{1}{1 + 0.75 + 0.6} = \frac{1}{\frac{235}{100}} = \frac{20}{47}\).
Hence, the required probability that it was the two headed coin is \(\frac{20}{47}\).

Question. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag. 
Answer: Let \(E_1, E_2\) and \(A\) denote the following events:
\(E_1\) = First bag is chosen, \(E_2\) = Second bag is chosen, and \(A\) = Two balls drawn at random are red.
Since, one of the bag is chosen at random:
\(\therefore P(E_1) = P(E_2) = \frac{1}{2}\).
If \(E_1\) has already occurred, i.e., first bag is chosen, the probability of drawing two red balls in this case is:
\(P\left(\frac{A}{E_1}\right) = \frac{^{5}C_2}{^{9}C_2} = \frac{10}{36}\).
Similarly, \(P\left(\frac{A}{E_2}\right) = \frac{^{3}C_2}{^{9}C_2} = \frac{3}{36}\).
We are required to find \(P\left(\frac{E_2}{A}\right)\).
By Bayes’ theorem,
\(P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)}\)
\(= \frac{\frac{1}{2} \times \frac{3}{36}}{\frac{1}{2} \times \frac{10}{36} + \frac{1}{2} \times \frac{3}{36}} = \frac{\frac{3}{72}}{\frac{10}{72} + \frac{3}{72}} = \frac{3}{13}\).
Therefore, required probability is \(\frac{3}{13}\).

Question. Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red, find the probability that two red balls were transferred from A to B.
Answer: Let us define the following events:
\(E_1\) = One red and one black ball is transferred,
\(E_2\) = Two red balls are transferred,
\(E_3\) = Two black balls are transferred,
and \(E\) = Drawn ball is red.
Then, \(P(E_1) = \frac{^{3}C_1 \times ^{5}C_1}{^{8}C_2} = \frac{3 \times 5}{28} = \frac{15}{28}\),
\(P(E_2) = \frac{^{3}C_2}{^{8}C_2} = \frac{3}{28}\), \(P(E_3) = \frac{^{5}C_2}{^{8}C_2} = \frac{10}{28}\).
\(P(E / E_1) = \frac{5}{10}\), \(P(E / E_2) = \frac{6}{10}\), \(P(E / E_3) = \frac{4}{10}\).
Now, required probability, \(P(E_2 / E)\):
\(P(E_2 / E) = \frac{P(E_2) \cdot P(E / E_2)}{P(E_1) \cdot P(E / E_1) + P(E_2) \cdot P(E / E_2) + P(E_3) \cdot P(E / E_3)}\)
\(= \frac{\frac{3}{28} \times \frac{6}{10}}{\frac{15}{28} \times \frac{5}{10} + \frac{3}{28} \times \frac{6}{10} + \frac{10}{28} \times \frac{4}{10}} = \frac{18}{75 + 18 + 40} = \frac{18}{133}\).

Question. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn at random from the bag and is found to be red. Find the probability that the balls are drawn from the first bag.
Answer: Let \(E_1\) = Event that Bag I is chosen, \(E_2\) = Event that Bag II is chosen, and \(A\) = Event that ball is drawn red.
Since, one of the bag is chosen at random:
\(\therefore P(E_1) = \frac{1}{2} = P(E_2)\).
Probability that ball is red and drawn from Bag I:
\(P\left(\frac{A}{E_1}\right) = \frac{4}{4+4} = \frac{1}{2}\).
Probability that ball is red and drawn from Bag II:
\(P\left(\frac{A}{E_2}\right) = \frac{2}{2+6} = \frac{1}{4}\).
By Bayes’ theorem,
\(P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \times P\left(\frac{A}{E_1}\right)}{P(E_1) \times P\left(\frac{A}{E_1}\right) + P(E_2) \times P\left(\frac{A}{E_2}\right)}\)
\(= \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} = \frac{2}{3}\).
Hence, the probability that the ball is drawn from Bag I, and the drawn ball is red = \(\frac{2}{3}\).

Question. By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
Answer: Let \(E_1\) be the event that person is suffering from TB and \(E_2\) be the event that person is not suffering from TB.
Let \(E\) be the event that the doctor diagnoses that person has TB.
Then, \(P(E_1) = \frac{1}{1000}\), \(P(E_2) = 1 - \frac{1}{1000} = \frac{999}{1000}\).
\(P\left(\frac{E}{E_1}\right) = P\text{(TB is detected when a person is actually suffering)} = 0.99 = \frac{990}{1000}\),
and \(P\left(\frac{E}{E_2}\right) = P\text{(TB is detected when a person is not actually suffering)} = 0.001 = \frac{1}{1000}\).
\(P\text{(A selected person has actually TB)}\), is given by:
\(P\left(\frac{E_1}{E}\right) = \frac{P(E_1) \cdot P\left(\frac{E}{E_1}\right)}{P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right)}\)
\(= \frac{\frac{1}{1000} \times \frac{990}{1000}}{\left(\frac{1}{1000} \times \frac{990}{1000}\right) + \left(\frac{999}{1000} \times \frac{1}{1000}\right)} = \frac{\frac{990}{1000000}}{\frac{990 + 999}{1000000}} = \frac{990}{1989} = \frac{110}{221}\).

Question. A laboratory blood test is 99% effective in detecting a certain disease, when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, then what is the probability that a person has disease, given that his test result is positive? 
Answer: Let \(E_1\) = Event that the person has disease, and \(E_2\) = Event that the person is healthy.
Then, \(P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001\) and \(P(E_2) = 1 - 0.001 = 0.999\).
Let \(E\): Event that test is positive.
\(P\left(\frac{E}{E_1}\right) = P\text{(result is positive, given that person has disease)} = 99\% = \frac{99}{100} = 0.99\).
\(P\left(\frac{E}{E_2}\right) = P\text{(result is positive, given that person does not have disease)} = 0.5\% = \frac{0.5}{100} = 0.005\).
\(\therefore\) Probability that a person has disease, given that test result is positive, is given by:
\(P\left(\frac{E_1}{E}\right) = \frac{P(E_1) \cdot P\left(\frac{E}{E_1}\right)}{P(E_1) \cdot P\left(\frac{E}{E_1}\right) + P(E_2) \cdot P\left(\frac{E}{E_2}\right)}\)
\(= \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985} = \frac{22}{133}\).

Question. There are four boxes, A, B, C and D, containing marbles. A contains 1 red, 6 white and 3 black marbles; B contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, then what is the probability that it was drawn from the Box A?
Answer: Let \(E_1, E_2, E_3\) and \(E_4\) be the events of selecting boxes, A, B, C and D, respectively.
Then, \(P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}\).
Let \(E\) = Event that the marble drawn is red. Then,
\(P(E/E_1) = \frac{1}{10}\), \(P(E/E_2) = \frac{6}{10} = \frac{3}{5}\), \(P(E/E_3) = \frac{8}{10} = \frac{4}{5}\), \(P(E/E_4) = 0\).
\(\therefore P(E_1/E) = \frac{P(E/E_1) \cdot P(E_1)}{P(E/E_1) \cdot P(E_1) + P(E/E_2) \cdot P(E_2) + P(E/E_3) \cdot P(E_3) + P(E/E_4) \cdot P(E_4)}\)
Ans. \(\frac{1}{15}\).

Question. Among the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual exams. At the end of year, one student is chosen at random from the college and he has A grade, what is the probability that the student is a hosteler? 
Answer: Let us define the events as:
\(E_1\): Students reside in a hostel,
\(E_2\): Students are day scholars,
and \(A\): Students get A grade.
Then, \(P(E_1) = \text{Probability that student reside in a hostel} = 60\% = \frac{60}{100}\),
and \(P(E_2) = \text{Probability that students are day scholars} = 1 - \frac{60}{100} = \frac{40}{100}\).
Also, \(P(A/E_1) = \text{Probability that hostelers get A grade} = 30\% = \frac{30}{100}\),
and \(P(A/E_2) = \text{Probability that students having day scholars get A grade} = 20\% = \frac{20}{100}\).
\(\dots\) The probability that the selecting student is a hosteler having A grade:
\(P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)}\) [by Bayes’ theorem]
\(= \frac{\frac{60}{100} \times \frac{30}{100}}{\left(\frac{60}{100} \times \frac{30}{100}\right) + \left(\frac{40}{100} \times \frac{20}{100}\right)} = \frac{1800}{1800 + 800} = \frac{18}{26} = \frac{9}{13}\).
Hence, required probability is \(\frac{9}{13}\).

Question. It is known that 20% of the students in a school have above 90% attendance and 80% of the students are irregular. Past year results show that 80% of students who have above 90% attendance and 20% of irregular students get A grade in their annual examination. At the end of a year, a student is chosen at random from the school and is found to have an ‘A’ grade. What is the probability that the student is irregular? 
Answer: Let us define the events as:
\(E_1\) = Event of students which have 90% attendance,
\(E_2\) = Event of students which are irregular,
and \(A\) = Event of students which attendance A grade.
Then, \(P(E_1) = 0.2\) and \(P(E_2) = 0.8\).
Also, \(P\left(\frac{A}{E_1}\right) = 0.8\) and \(P\left(\frac{A}{E_2}\right) = 0.2\).
Using the law of total probability, we compute \(P(A)\):
\(P(A) = P\left(\frac{A}{E_1}\right)P(E_1) + P\left(\frac{A}{E_2}\right)P(E_2)\)
\(\Rightarrow P(A) = (0.8)(0.2) + (0.2)(0.8) \Rightarrow P(A) = 0.16 + 0.16 \Rightarrow P(A) = 0.32\).
Now, using Bayes’ theorem,
\(P\left(\frac{E_2}{A}\right) = \frac{P\left(\frac{A}{E_2}\right) P(E_2)}{P(A)} = \frac{(0.2)(0.8)}{0.32} = 0.5\).
The probability that the student is irregular given they have an A grade is 0.5 or 50%.

Question. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time, a patient can choose anyone of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga. Interpret the result and state which of the above stated methods, is more beneficial for the patient?
Answer: Let \(E_1\): The patient follows meditation and yoga,
\(E_2\): The patient uses drug.
Then, \(E_1\) and \(E_2\) are mutually exclusive and \(P(E_1) = P(E_2) = 1/2\).
Also, let \(E\): The selected patient suffers a heart attack.
Then, \(P(E/E_1) = \frac{40}{100} \left(1 - \frac{30}{100}\right) = \frac{40}{100} \times \frac{70}{100} = \frac{28}{100}\),
and \(P(E/E_2) = \frac{40}{100} \left(1 - \frac{25}{100}\right) = \frac{40}{100} \times \frac{75}{100} = \frac{30}{100}\).
\(\therefore\) P (patient who suffers heart attack follows meditation and yoga) = \(P(E_1/E)\):
\(P(E_1/E) = \frac{P(E/E_1) \cdot P(E_1)}{P(E/E_1) \cdot P(E_1) + P(E/E_2) \cdot P(E_2)}\) [using Bayes’ theorem]
\(= \frac{\frac{28}{100} \times \frac{1}{2}}{\left(\frac{28}{100} \times \frac{1}{2}\right) + \left(\frac{30}{100} \times \frac{1}{2}\right)} = \frac{28}{58} = \frac{14}{29}\).
Hence, Yoga course and meditation are more beneficial for the heart patient.