NCERT Solutions Class-12-Chapter-13-Probability
Excercise-13.3
Note:
Theorem of Total Probability: Let \( \{E_1, E_2, \dots, E_n\} \) be a partition of the sample space S. Then for any event A,
\( P(A) = \sum_{j=1}^{n} P(E_j) P(A|E_j) \).
Bayes' Theorem: If \( E_1, E_2, \dots, E_n \) are mutually exclusive and exhaustive events, then for any event A:
\( P(E_i|A) = \frac{P(E_i) P(A|E_i)}{\sum_{j=1}^{n} P(E_j) P(A|E_j)} \).
Q1
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution:
Let \( R_1 \) be the event that the first ball is red, and \( B_1 \) be the event that the first ball is black. Let \( R_2 \) be the event that the second ball is red.
\( P(R_1) = \frac{5}{10} = \frac{1}{2} \), \( P(B_1) = \frac{5}{10} = \frac{1}{2} \).
Case 1: First ball is Red.
Urn now has \( 5+2=7 \) Red and 5 Black balls (Total 12).
\( P(R_2|R_1) = \frac{7}{12} \).
Case 2: First ball is Black.
Urn now has 5 Red and \( 5+2=7 \) Black balls (Total 12).
\( P(R_2|B_1) = \frac{5}{12} \).
By Total Probability Theorem:
\( P(R_2) = P(R_1)P(R_2|R_1) + P(B_1)P(R_2|B_1) \)
\( = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2} \).
Q2
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 from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution:
Let \( E_1 \) be the event of choosing Bag I and \( E_2 \) be the event of choosing Bag II.
\( P(E_1) = P(E_2) = \frac{1}{2} \).
Let A be the event that the ball drawn is Red.
Bag I: 4R, 4B. \( P(A|E_1) = \frac{4}{8} = \frac{1}{2} \).
Bag II: 2R, 6B. \( P(A|E_2) = \frac{2}{8} = \frac{1}{4} \).
Using Bayes' Theorem, we need \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3} \).
Q3
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars. Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade. At the end of the year, one student is chosen at random and has an A grade, what is the probability that the student is a hostlier?
Solution:
Let \( E_1 \): Student is a hostlier, \( E_2 \): Student is a day scholar.
\( P(E_1) = 0.6 \), \( P(E_2) = 0.4 \).
Let A: Student attains Grade A.
\( P(A|E_1) = 0.3 \), \( P(A|E_2) = 0.2 \).
Using Bayes' Theorem for \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26} = \frac{18}{26} = \frac{9}{13} \).
Q4
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \( \frac{3}{4} \) be the probability that he knows the answer and \( \frac{1}{4} \) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \( \frac{1}{4} \). What is the probability that the student knows the answer given that he answered it correctly?
Solution:
Let \( E_1 \): Student knows the answer, \( E_2 \): Student guesses.
\( P(E_1) = \frac{3}{4} \), \( P(E_2) = \frac{1}{4} \).
Let A: Answer is correct.
If student knows the answer, probability of correct answer is 1. So, \( P(A|E_1) = 1 \).
If student guesses, \( P(A|E_2) = \frac{1}{4} \).
Using Bayes' Theorem:
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{12}{16}}{\frac{12+1}{16}} = \frac{12}{13} \).
Q5
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. If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution:
Let \( E_1 \): Person has disease, \( E_2 \): Person does not have disease.
\( P(E_1) = 0.1\% = 0.001 \).
\( P(E_2) = 1 - 0.001 = 0.999 \).
Let A: Test result is positive.
\( P(A|E_1) \) (True Positive) = 99% = 0.99.
\( P(A|E_2) \) (False Positive) = 0.5% = 0.005.
Using Bayes' Theorem:
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} \)
\( = \frac{0.00099}{0.00099 + 0.004995} = \frac{990}{990 + 4995} = \frac{990}{5985} = \frac{22}{133} \).
Q6
There are three coins. One is a two headed coin, another is a biased coin that comes up heads 75% of the time 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?
Solution:
Let \( E_1 \): Two headed coin, \( E_2 \): Biased coin, \( E_3 \): Unbiased coin.
\( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
Let A: Coin shows Head.
\( P(A|E_1) = 1 \) (Two headed).
\( P(A|E_2) = 75\% = \frac{3}{4} \).
\( P(A|E_3) = 50\% = \frac{1}{2} \).
Using Bayes' Theorem for \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)} \)
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3}(1 + \frac{3}{4} + \frac{1}{2})} = \frac{1}{\frac{4+3+2}{4}} = \frac{4}{9} \).
Q7
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
Total drivers = \( 2000 + 4000 + 6000 = 12000 \).
\( P(E_1) \) (Scooter) = \( \frac{2000}{12000} = \frac{1}{6} \).
\( P(E_2) \) (Car) = \( \frac{4000}{12000} = \frac{1}{3} \).
\( P(E_3) \) (Truck) = \( \frac{6000}{12000} = \frac{1}{2} \).
Probabilities of accident (A):
\( P(A|E_1) = 0.01 \), \( P(A|E_2) = 0.03 \), \( P(A|E_3) = 0.15 \).
\( P(E_1|A) = \frac{\frac{1}{6}(0.01)}{\frac{1}{6}(0.01) + \frac{2}{6}(0.03) + \frac{3}{6}(0.15)} \) (using common denominator 6)
\( = \frac{0.01}{0.01 + 0.06 + 0.45} = \frac{0.01}{0.52} = \frac{1}{52} \).
Q8
A factory has two machines A and B. Machine A produced 60% of the items and machine B produced 40%. Further, 2% of items from A and 1% from B were defective. All items are put into one stockpile and one item is chosen at random and found to be defective. What is the probability that it was produced by machine B?
Solution:
\( E_1 \): Machine A, \( E_2 \): Machine B.
\( P(E_1) = 0.6 \), \( P(E_2) = 0.4 \).
A: Defective item.
\( P(A|E_1) = 0.02 \), \( P(A|E_2) = 0.01 \).
Required \( P(E_2|A) \):
\( = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.012 + 0.004} = \frac{0.004}{0.016} = \frac{4}{16} = \frac{1}{4} \).
Q9
Two groups are competing for the position on the Board of directors. The probabilities that the first and the second groups 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.
Solution:
\( E_1 \): Group 1 wins (\( 0.6 \)), \( E_2 \): Group 2 wins (\( 0.4 \)).
A: New product introduced.
\( P(A|E_1) = 0.7 \), \( P(A|E_2) = 0.3 \).
Required \( P(E_2|A) \):
\( = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9} \).
Q10
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, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution:
\( E_1 \): Die shows 5 or 6 (\( P(E_1) = \frac{2}{6} = \frac{1}{3} \)).
\( E_2 \): Die shows 1, 2, 3 or 4 (\( P(E_2) = \frac{4}{6} = \frac{2}{3} \)).
A: Exactly one head obtained.
For \( E_1 \) (3 tosses): P(1 Head) = \( \binom{3}{1}(\frac{1}{2})^1(\frac{1}{2})^2 = \frac{3}{8} \).
For \( E_2 \) (1 toss): P(1 Head) = \( \frac{1}{2} \).
Required \( P(E_2|A) \):
\( = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{3+8}{24}} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11} \).
Q11
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Solution:
\( P(E_1) \) (A) = 0.50, \( P(E_2) \) (B) = 0.30, \( P(E_3) \) (C) = 0.20.
Defective (D):
\( P(D|E_1) = 0.01 \), \( P(D|E_2) = 0.05 \), \( P(D|E_3) = 0.07 \).
Required \( P(E_1|D) \):
\( = \frac{0.5 \times 0.01}{0.5(0.01) + 0.3(0.05) + 0.2(0.07)} = \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34} \).
Q12
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Solution:
Let \( E_1 \): Lost card is Diamond, \( E_2 \): Lost card is not Diamond.
\( P(E_1) = \frac{13}{52} = \frac{1}{4} \), \( P(E_2) = \frac{39}{52} = \frac{3}{4} \).
A: Two cards drawn are diamonds.
If \( E_1 \): 12 Diamonds left in 51. \( P(A|E_1) = \frac{^{12}C_2}{^{51}C_2} \).
If \( E_2 \): 13 Diamonds left in 51. \( P(A|E_2) = \frac{^{13}C_2}{^{51}C_2} \).
Required \( P(E_1|A) \):
\( = \frac{\frac{1}{4} \binom{12}{2}}{\frac{1}{4} \binom{12}{2} + \frac{3}{4} \binom{13}{2}} \) (Common denominator \( 4 \times \binom{51}{2} \) cancels out)
\( = \frac{1 \times 66}{1 \times 66 + 3 \times 78} = \frac{66}{66 + 234} = \frac{66}{300} = \frac{11}{50} \).
Q13
Probability that A speaks truth is \( \frac{4}{5} \). A coin is tossed. A reports that a head appears. The probability that actually there was head is:
(A) \( \frac{4}{5} \)
(B) \( \frac{1}{2} \)
(C) \( \frac{1}{5} \)
(D) \( \frac{2}{5} \)
(A) \( \frac{4}{5} \)
(B) \( \frac{1}{2} \)
(C) \( \frac{1}{5} \)
(D) \( \frac{2}{5} \)
Answer: (A)
Solution:
Let \( E_1 \): Head occurs (\( P(E_1) = \frac{1}{2} \)).
Let \( E_2 \): Tail occurs (\( P(E_2) = \frac{1}{2} \)).
Let A: A reports that a Head appears.
- \( P(A|E_1) \): A reports Head given Head occurred (Truth). \( P(A|E_1) = \frac{4}{5} \).
- \( P(A|E_2) \): A reports Head given Tail occurred (Lie). \( P(A|E_2) = 1 - \frac{4}{5} = \frac{1}{5} \).
We need to find the probability that there was actually a head, given A reports a head, i.e., \( P(E_1|A) \).
Using Bayes' Theorem:
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \]
\[ P(E_1|A) = \frac{\frac{1}{2} \times \frac{4}{5}}{\left(\frac{1}{2} \times \frac{4}{5}\right) + \left(\frac{1}{2} \times \frac{1}{5}\right)} \]
\[ P(E_1|A) = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{1}{10}} = \frac{4}{5} \]
Thus, the correct option is (A).
Q14
If A and B are two events such that \( A \subset B \) and \( P(B) \neq 0 \), then which of the following is correct?
(A) \( P(A|B) = \frac{P(B)}{P(A)} \)
(B) \( P(A|B) < P(A) \)
(C) \( P(A|B) \ge P(A) \)
(D) None of these
(A) \( P(A|B) = \frac{P(B)}{P(A)} \)
(B) \( P(A|B) < P(A) \)
(C) \( P(A|B) \ge P(A) \)
(D) None of these
Answer: (C)
Solution:
We are given that \( A \subset B \). This implies that the intersection of A and B is just A itself, i.e., \( A \cap B = A \).
Using the formula for conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \]
We know that for any event B, the probability \( P(B) \) lies in the range \( 0 < P(B) \le 1 \) (since \( P(B) \neq 0 \)).
If we divide \( P(A) \) by a number \( P(B) \) which is less than or equal to 1, the result will be greater than or equal to \( P(A) \).
\[ \frac{P(A)}{P(B)} \ge P(A) \quad (\text{since } P(B) \le 1) \]
Therefore, \( P(A|B) \ge P(A) \).