CBSE Class 12 Mathematics Probability Notes Set 04

Important Notes and Formulas

Probability of Independent Events (Addition Theorem)

Notations: \( P(A \cup B) \) = Probability of happening of A or B = Probability of occurrence of at least one event A or B; \( P(A \cap B) \) = Probability of happening of A and B together.
When events are independent: If A and B are independent events, then \( P(A \cap B) = P(A).P(B) \)
∴ \( P(A \cup B) = P(A) + P(B) - P(A).P(B) \).

Conditional Probability

Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A under the condition that B has already occurred and P(B) ≠ 0, is called the conditional probability and it is denoted by P(A/B).
Thus, P(A/B) = Probability of occurrence of A, given that B has already happened.
\( = \frac{P(A \cap B)}{P(B)} = \frac{n(A \cap B)}{n(B)} \).
Similarly, P(B/A) = Probability of occurrence of B, given that A has already happened.
\( = \frac{P(A \cap B)}{P(A)} = \frac{n(A \cap B)}{n(A)} \).
Note: Sometimes, P(A/B) is also used to denote the probability of occurrence of A when B occurs. Similarly, P(B/A) is used to denote the probability of occurrence of B when A occurs.

(1) Multiplication theorems on probability
(i) If A and B are two events associated with a random experiment, then \( P(A \cap B) = P(A).P(B/A) \), if P(A) ≠ 0 or \( P(A \cap B) = P(B).P(A/B) \), if P(B) ≠ 0.
(ii) Extension of multiplication theorem: If \( A_1, A_2, \ldots, A_n \) are n events related to a random experiment, then \( P(A_1 \cap A_2 \cap A_3 \cap \ldots \cap A_n) = P(A_1)P(A_2/A_1)P(A_3/A_1 \cap A_2) \ldots P(A_n/A_1 \cap A_2 \cap \ldots \cap A_{n-1}) \),
where \( P(A_i/A_1 \cap A_2 \cap \ldots \cap A_{i-1}) \) represents the conditional probability of the event \( A_i \), given that the events \( A_1, A_2, \ldots, A_{i-1} \) have already happened.
(iii) Multiplication theorems for independent events: If A and B are independent events associated with a random experiment, then \( P(A \cap B) = P(A).P(B) \) i.e., the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities.
By multiplication theorem, we have \( P(A \cap B) = P(A).P(B/A) \).
Since A and B are independent events, therefore \( P(B/A) = P(B) \). Hence, \( P(A \cap B) = P(A).P(B) \).
(iv) Extension of multiplication theorem for independent events: If \( A_1, A_2, \ldots, A_n \) are independent events associated with a random experiment, then \( P(A_1 \cap A_2 \cap A_3 \cap \ldots \cap A_n) = P(A_1)P(A_2) \ldots P(A_n) \).
By multiplication theorem, we have
\( P(A_1 \cap A_2 \cap A_3 \cap \ldots \cap A_n) = P(A_1)P(A_2/A_1)P(A_3/A_1 \cap A_2) \ldots P(A_n/A_1 \cap A_2 \cap \ldots \cap A_{n-1}) \)
Since \( A_1, A_2, \ldots, A_{n-1}, A_n \) are independent events, therefore
\( P(A_2/A_1) = P(A_2),\ P(A_3/A_1 \cap A_2) = P(A_3),\ \ldots,\ P(A_n/A_1 \cap A_2 \cap \ldots \cap A_{n-1}) = P(A_n) \)
Hence, \( P(A_1 \cap A_2 \cap \ldots \cap A_n) = P(A_1)P(A_2) \ldots P(A_n) \).

(2) Probability of at least one of the n independent events
If \( p_1, p_2, p_3, \ldots, p_n \) be the probabilities of happening of n independent events \( A_1, A_2, A_3, \ldots, A_n \) respectively, then
(i) Probability of happening none of them
\( = P(\bar{A}_1 \cap \bar{A}_2 \cap \bar{A}_3 \cap \ldots \cap \bar{A}_n) = P(\bar{A}_1).P(\bar{A}_2).P(\bar{A}_3) \ldots P(\bar{A}_n) = (1 - p_1)(1 - p_2)(1 - p_3) \ldots (1 - p_n) \).
(ii) Probability of happening at least one of them
\( = P(A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_n) = 1 - P(\bar{A}_1)P(\bar{A}_2)P(\bar{A}_3) \ldots P(\bar{A}_n) = 1 - (1 - p_1)(1 - p_2)(1 - p_3) \ldots (1 - p_n) \).
(iii) Probability of happening of first event and not happening of the remaining
\( = P(A_1)P(\bar{A}_2)P(\bar{A}_3) \ldots P(\bar{A}_n) = p_1(1 - p_2)(1 - p_3) \ldots (1 - p_n) \)

Total Probability and Bayes' Rule

(1) The law of total probability: Let S be the sample space and let \( E_1, E_2, \ldots, E_n \) be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with \( E_1 \) or \( E_2 \) or … or \( E_n \), then \( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + \ldots + P(E_n)P(A/E_n) \).

(2) Bayes' rule: Let S be a sample space and \( E_1, E_2, \ldots, E_n \) be n mutually exclusive events such that \( \bigcup_{i=1}^{n} E_i = S \) and \( P(E_i) > 0 \) for i = 1, 2, …, n. We can think of \( E_i \)'s as the causes that lead to the outcome of an experiment. The probabilities \( P(E_i) \), i = 1, 2, …, n are called prior probabilities. Suppose the experiment results in an outcome of event A, where P(A) > 0. We have to find the probability that the observed event A was due to cause \( E_i \), that is, we seek the conditional probability \( P(E_i/A) \). These probabilities are called posterior probabilities, given by Bayes' rule as
\( P(E_i/A) = \frac{P(E_i).P(A/E_i)}{\sum_{k=1}^{n} P(E_k)P(A/E_k)} \).

Solved Examples

Example 1. The probability that a man will be alive in 20 years is \( \frac{3}{5} \) and the probability that his wife will be alive in 20 years is \( \frac{2}{3} \). Then the probability that at least one will be alive in 20 years is [Bihar CEE 1994]
(a) \( \frac{13}{15} \)
(b) \( \frac{7}{15} \)
(c) \( \frac{4}{15} \)
(d) None of these
Answer:
Ans. (a) \( \frac{13}{15} \)
Let A be the event that the husband will be alive 20 years. B be the event that the wife will be alive 20 years. Clearly A and B are independent events. ∴ \( P(A \cap B) = P(A)P(B) \).
Given \( P(A) = \frac{3}{5},\ P(B) = \frac{2}{3} \).
The probability that at least one of them will be alive 20 years is
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(A).P(B) = \frac{3}{5} + \frac{2}{3} - \frac{3}{5} \cdot \frac{2}{3} = \frac{9 + 10 - 6}{15} = \frac{13}{15} \).

Example 2. Let A and B be two events such that \( P(A) = 0.3 \) and \( P(A \cup B) = 0.8 \). If A and B are independent events, then P(B) = [IIT 1990; UPSEAT 2001, 02]
(a) \( \frac{5}{6} \)
(b) \( \frac{5}{7} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{2}{5} \)
Answer:
Ans. (b) \( \frac{5}{7} \)
Here \( P(A \cup B) = 0.8 \), P(A) = 0.3 and A and B are independent events.
Let P(B) = x. ∴ \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow P(A \cup B) = P(A) + P(B) - P(A).P(B) \)
⇒ \( 0.8 = 0.3 + x - 0.3x \Rightarrow x = \frac{5}{7} \).

Example 3. The probability that A speaks truth is \( \frac{4}{5} \), while this probability for B is \( \frac{3}{4} \). The probability that they contradict each other when asked to speak on a fact is [AIEEE 2004]
(a) \( \frac{4}{5} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{7}{20} \)
(d) \( \frac{3}{20} \)
Answer:
Ans. (c) \( \frac{7}{20} \)
Let E be the event that B speaks truth and F be the event that A speaks truth.
Now \( P(E) = \frac{75}{100} = \frac{3}{4} \) and \( P(F) = \frac{80}{100} = \frac{4}{5} \).
∴ P (A and B contradict each other)
= P [(B tells truth and A tells lie) or (B tells lie and A tells truth)]
\( = P[(E \cap \bar{F}) \cup (\bar{E} \cap F)] = P(E).P(\bar{F}) + P(\bar{E}).P(F) = \frac{3}{4} \times \frac{1}{5} + \frac{1}{4} \times \frac{4}{5} = \frac{7}{20} \).

Example 4. A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II, III are p, q and \( \frac{1}{2} \) respectively. If the probability that the student is successful is \( \frac{1}{2} \), then [IIT 1986]
(a) p = 1, q = 0
(b) \( p = \frac{2}{3},\ q = \frac{1}{2} \)
(c) There are infinitely many values of p and q
(d) All of the above
Answer:
Ans. (d) All of the above
Let A, B and C be the events that the student is successful in test I, II and III respectively, then P (the student is successful)
\( = P[(A \cap B \cap C') \cup (A \cap B' \cap C) \cup (A \cap B \cap C)] = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A \cap B \cap C) \)
\( = P(A).P(B).P(C') + P(A).P(B').P(C) + P(A).P(B).P(C) \) [∵ A, B, C are independent]
\( = pq\left(1 - \frac{1}{2}\right) + p(1 - q)\left(\frac{1}{2}\right) + pq\left(\frac{1}{2}\right) = \frac{1}{2}p(1 + q) \Rightarrow \frac{1}{2} = \frac{1}{2}p(1 + q) \Rightarrow p(1 + q) = 1 \).
This equation has infinitely many values of p and q. Also p = 1, q = 0 and \( p = \frac{2}{3},\ q = \frac{1}{2} \) both satisfy p(1 + q) = 1, so all of the above are true.

Example 5. A man and his wife appear for an interview for two posts. The probability of the husband's selection is \( \frac{1}{7} \) and that of wife's selection is \( \frac{1}{5} \). What is the probability that only one of them will be selected? [AISSE]
(a) \( \frac{1}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{7} \)
(d) None of these
Answer:
Ans. (b) \( \frac{2}{7} \)
The probability of husband is not selected \( = 1 - \frac{1}{7} = \frac{6}{7} \); the probability that wife is not selected \( = 1 - \frac{1}{5} = \frac{4}{5} \)
The probability that only husband is selected \( = \frac{1}{7} \times \frac{4}{5} = \frac{4}{35} \); the probability that only wife is selected \( = \frac{1}{5} \times \frac{6}{7} = \frac{6}{35} \)
Hence, required probability \( = \frac{6}{35} + \frac{4}{35} = \frac{10}{35} = \frac{2}{7} \).

Example 6. A purse contains 4 copper coins and 3 silver coins, the second purse contains 6 copper coins and 2 silver coins. If a coin is drawn out of any purse, then the probability that it is a copper coin is [Ranchi BIT]
(a) 4/7
(b) 3/4
(c) 37/56
(d) None of these
Answer:
Ans. (c) 37/56
Required probability \( = \frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{6}{8} = \frac{37}{56} \).

Example 7. If \( 4P(A) = 6P(B) = 10P(A \cap B) = 1 \), then \( P\left(\frac{B}{A}\right) = \) [MP PET 2003]
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{7}{10} \)
(d) \( \frac{19}{60} \)
Answer:
Ans. (a) \( \frac{2}{5} \)
\( P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)} = \frac{(1/10)}{(1/4)} = \frac{2}{5} \).

Example 8. A coin is tossed three times in succession. If E is the event that there are at least two heads and F is the event in which first throw is a head, then \( P\left(\frac{E}{F}\right) = \) [MP PET 1996]
(a) \( \frac{3}{4} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{8} \)
Answer:
Ans. (a) \( \frac{3}{4} \)
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(E) = 4, n(F) = 4 and \( n(E \cap F) = 3 \)
∴ \( P\left(\frac{E}{F}\right) = \frac{P(E \cap F)}{P(F)} = \frac{3/8}{4/8} = \frac{3}{4} \).

Example 9. Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second an honour card is (before drawing second card first card is not placed again in the pack) [UPSEAT]
(a) 1/26
(b) 5/52
(c) 5/221
(d) 4/13
Answer:
Ans. (c) 5/221
\( P(E_1) = \frac{4}{52} = \frac{1}{13},\ P\left(\frac{E_2}{E_1}\right) = \frac{15}{51} = \frac{5}{17} \)
\( P(E_1 \cap E_2) = P(E_1).P\left(\frac{E_2}{E_1}\right) = \frac{1}{13} \cdot \frac{5}{17} = \frac{5}{221} \).

Example 10. If A and B are two events such that P(A) ≠ 0 and P(B) ≠ 1, then \( P\left(\frac{\bar{A}}{\bar{B}}\right) = \) [IIT 1982; RPET 1995, 2000; DCE 2000; UPSEAT 2001]
(a) \( 1 - P\left(\frac{A}{B}\right) \)
(b) \( 1 - P\left(\frac{\bar{A}}{B}\right) \)
(c) \( \frac{1 - P(A \cup B)}{P(\bar{B})} \)
(d) \( \frac{P(\bar{A})}{P(\bar{B})} \)
Answer:
Ans. (c) \( \frac{1 - P(A \cup B)}{P(\bar{B})} \)
\( P\left(\frac{\bar{A}}{\bar{B}}\right) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{P(\overline{A \cup B})}{P(\bar{B})} = \frac{1 - P(A \cup B)}{P(\bar{B})} \).

Example 11. If A and B are two events such that \( P(A \cup B) = P(A \cap B) \), then the true relation is [IIT 1985]
(a) \( P(A) + P(B) = 0 \)
(b) \( P(A) + P(B) = P(A)P\left(\frac{B}{A}\right) \)
(c) \( P(A) + P(B) = 2P(A)P\left(\frac{B}{A}\right) \)
(d) None of these
Answer:
Ans. (c) \( P(A) + P(B) = 2P(A)P\left(\frac{B}{A}\right) \)
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow P(A \cap B) = P(A) + P(B) - P(A \cap B) \) {∵ \( P(A \cap B) = P(A \cup B) \)}
⇒ \( 2P(A \cap B) = P(A) + P(B) \Rightarrow 2P(A).\frac{P(A \cap B)}{P(A)} = P(A) + P(B) \Rightarrow 2P(A)P\left(\frac{B}{A}\right) = P(A) + P(B) \).

Example 12. Let E and F be two independent events. The probability that both E and F happens is \( \frac{1}{12} \) and the probability that neither E nor F happens is \( \frac{1}{2} \), then [IIT 1993]
(a) \( P(E) = \frac{1}{3},\ P(F) = \frac{1}{4} \)
(b) \( P(E) = \frac{1}{2},\ P(F) = \frac{1}{6} \)
(c) \( P(E) = \frac{1}{6},\ P(F) = \frac{1}{2} \)
(d) None of these
Answer:
Ans. (a) \( P(E) = \frac{1}{3},\ P(F) = \frac{1}{4} \)
We are given \( P(E \cap F) = \frac{1}{12} \) and \( P(\bar{E} \cap \bar{F}) = \frac{1}{2} \)
⇒ \( P(E).P(F) = \frac{1}{12} \) …(i) and \( P(\bar{E}).P(\bar{F}) = \frac{1}{2} \) …(ii)
⇒ \( \{1 - P(E)\}\{1 - P(F)\} = \frac{1}{2} \Rightarrow 1 + P(E)P(F) - P(E) - P(F) = \frac{1}{2} \Rightarrow 1 + \frac{1}{12} - [P(E) + P(F)] = \frac{1}{2} \)
⇒ \( P(E) + P(F) = \frac{7}{12} \) …(iii)
On solving (i) and (iii), we get \( P(E) = \frac{1}{3}, \frac{1}{4} \) and \( P(F) = \frac{1}{4}, \frac{1}{3} \).

Example 13. Let p denotes the probability that a man aged x years will die in a year. The probability that out of n men \( A_1, A_2, A_3, \ldots, A_n \) each aged x, \( A_1 \) will die in a year and will be the first to die, is [MNR 1987; UPSEAT 2000]
(a) \( \frac{1}{n}[1 - (1 - p)^n] \)
(b) \( [1 - (1 - p)^n] \)
(c) \( \frac{1}{n - 1}[1 - (1 - p)^n] \)
(d) None of these
Answer:
Ans. (a) \( \frac{1}{n}[1 - (1 - p)^n] \)
Let \( E_i \) denotes the event that \( A_i \) dies in a year.
Then \( P(E_i) = p \) and \( P(E_i') = 1 - p \) for i = 1, 2, …, n
P (none of \( A_1, A_2, \ldots, A_n \) dies in a year) \( = P(E_1' \cap E_2' \cap \ldots \cap E_n') = P(E_1')P(E_2') \ldots P(E_n') = (1 - p)^n \), because \( E_1, E_2, \ldots, E_n \) are independent.
Let E denote the event that at least one of \( A_1, A_2, \ldots, A_n \) dies in a year.
Then \( P(E) = 1 - P(E_1' \cap E_2' \cap \ldots \cap E_n') = 1 - (1 - p)^n \)
Let F denote the event that \( A_1 \) is the first to die.
Then \( P(F/E) = \frac{1}{n} \). Also, \( P(F) = P(E).P(F/E) = \frac{1}{n}[1 - (1 - p)^n] \).

Example 14. A problem of mathematics is given to three students whose chances of solving the problem are 1/3, 1/4 and 1/5 respectively. The probability that the question will be solved is [BIT Ranchi 1991; MP PET 1990]
(a) 2/3
(b) 3/4
(c) 4/5
(d) 3/5
Answer:
Ans. (d) 3/5
The probabilities of students not solving the problem are \( 1 - \frac{1}{3} = \frac{2}{3},\ 1 - \frac{1}{4} = \frac{3}{4} \) and \( 1 - \frac{1}{5} = \frac{4}{5} \).
Therefore the probability that the problem is not solved by any one of them \( = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5} \).
Hence, the probability that problem is solved \( = 1 - \frac{2}{5} = \frac{3}{5} \).

Example 15. The probability of happening an event A in one trial is 0.4. The probability that the event A happens at least once in three independent trials is [IIT 1980; Kurukshetra CEE 1998; DCE 2001]
(a) 0.936
(b) 0.784
(c) 0.904
(d) 0.216
Answer:
Ans. (b) 0.784
Here P(A) = 0.4 and \( P(\bar{A}) = 0.6 \)
Probability that A does not happen at all \( = (0.6)^3 \). Thus required probability \( = 1 - (0.6)^3 = 0.784 \).

Example 16. In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. Then the probability that the bolt drawn is defective is
(a) 0.0345
(b) 0.345
(c) 3.45
(d) 0.0034
Answer:
Ans. (a) 0.0345
Let \( E_1, E_2, E_3 \) and A be the events defined as follows:
\( E_1 \) = the bolt is manufactured by machine A; \( E_2 \) = the bolt is manufactured by machine B; \( E_3 \) = the bolt is manufactured by machine C, and A = the bolt is defective.
Then \( P(E_1) = \frac{25}{100} = \frac{1}{4},\ P(E_2) = \frac{35}{100},\ P(E_3) = \frac{40}{100} \).
\( P(A/E_1) \) = Probability that the bolt drawn is defective given the condition that it is manufactured by machine A = 5/100.
Similarly \( P(A/E_2) = \frac{4}{100} \) and \( P(A/E_3) = \frac{2}{100} \).
Using the law of total probability, we have \( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3) \)
\( = \frac{25}{100} \times \frac{5}{100} + \frac{35}{100} \times \frac{4}{100} + \frac{40}{100} \times \frac{2}{100} = 0.0345 \).

Example 17. A lot contains 20 articles. The probability that the lot contains 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn at random one by one without replacement and tested till all the defective articles are found. The probability that the testing procedure ends at the twelfth testing is
(a) \( \frac{9}{1900} \)
(b) \( \frac{19}{1000} \)
(c) \( \frac{99}{1900} \)
(d) \( \frac{19}{900} \)
Answer:
Ans. (c) \( \frac{99}{1900} \)
The testing procedure may terminate at the twelfth testing in two mutually exclusive ways.
(I) When lot contains 2 defective articles. (II) When lot contains 3 defective articles.
Consider the following events.
A = Testing procedure ends at the twelfth testing.
\( A_1 \) = Lot contains 2 defective articles.
\( A_2 \) = Lot contains 3 defective articles.
Required probability
\( = P(A) = P[(A \cap A_1) \cup (A \cap A_2)] = P(A \cap A_1) + P(A \cap A_2) = P(A_1)P(A/A_1) + P(A_2)P(A/A_2) \)
Now, \( P(A/A_1) \) = Probability that first 11 draws contain 10 non-defective and one defective and 12th draw contains a defective article \( = \frac{^{18}C_{10} \times {^{2}C_{1}}}{^{20}C_{11}} \times \frac{1}{9} \).
And \( P(A/A_2) \) = Probability that first 11 draws contain 9 non-defective and 2 defective articles and 12th draw contains a defective article \( = \frac{^{17}C_{9} \times {^{3}C_{2}}}{^{20}C_{11}} \times \frac{1}{9} \)
Hence, required probability \( = 0.4 \times \frac{^{18}C_{10} \times {^{2}C_{1}}}{^{20}C_{11}} \times \frac{1}{9} + 0.6 \times \frac{^{17}C_{9} \times {^{3}C_{2}}}{^{20}C_{11}} \times \frac{1}{9} = \frac{99}{1900} \).

Example 18. A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from B is [BIT Ranchi 1988; IIT 1976]
(a) \( \frac{5}{14} \)
(b) \( \frac{5}{16} \)
(c) \( \frac{5}{18} \)
(d) \( \frac{25}{52} \)
Answer:
Ans. (d) \( \frac{25}{52} \)
Let \( E_1 \) be the event that the ball is drawn from bag A, \( E_2 \) the event that it is drawn from bag B and E that the ball is red.
We have to find \( P(E_2/E) \).
Since both the bags are equally likely to be selected,
we have \( P(E_1) = P(E_2) = \frac{1}{2} \). Also \( P(E/E_1) = 3/5 \) and \( P(E/E_2) = 5/9 \).
Hence by Bayes' theorem, we have \( P(E_2/E) = \frac{P(E_2)P(E/E_2)}{P(E_1)P(E/E_1) + P(E_2)P(E/E_2)} = \frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{5}{9}} = \frac{25}{52} \).

Example 19. A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is
(a) \( \frac{3}{8} \)
(b) \( \frac{1}{5} \)
(c) \( \frac{3}{4} \)
(d) None of these
Answer:
Ans. (a) \( \frac{3}{8} \)
Let E denote the event that a six occurs and A the event that the man reports that it is a '6', we have
\( P(E) = \frac{1}{6},\ P(E') = \frac{5}{6},\ P(A/E) = \frac{3}{4} \) and \( P(A/E') = \frac{1}{4} \).
By Bayes' theorem, \( P(E/A) = \frac{P(E).P(A/E)}{P(E).P(A/E) + P(E')P(A/E')} = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{3}{8} \).

Example 20. A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability that the missing card is black, is
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{^{25}C_{13}}{^{51}C_{13}} \)
Answer:
Ans. (b) \( \frac{2}{3} \)
Let \( A_1 \) be the event that the black card is lost, \( A_2 \) be the event that the red card is lost and let E be the event that first 13 cards examined are red.
Then the required probability \( = P\left(\frac{A_1}{E}\right) \). We have \( P(A_1) = P(A_2) = \frac{1}{2} \); as black and red cards were initially equal in number.
Also \( P\left(\frac{E}{A_1}\right) = \frac{^{26}C_{13}}{^{51}C_{13}} \) and \( P\left(\frac{E}{A_2}\right) = \frac{^{25}C_{13}}{^{51}C_{13}} \).
The required probability \( = P\left(\frac{A_1}{E}\right) = \frac{P(E/A_1)P(A_1)}{P(E/A_1)P(A_1) + P(E/A_2)P(A_2)} = \frac{\frac{1}{2} \cdot \frac{^{26}C_{13}}{^{51}C_{13}}}{\frac{1}{2} \cdot \frac{^{26}C_{13}}{^{51}C_{13}} + \frac{1}{2} \cdot \frac{^{25}C_{13}}{^{51}C_{13}}} = \frac{2}{3} \).