Chapter 13 PROBABILITY EXERCISE 13.3 solution

NCERT Solutions for Class 12 Maths – Chapter 13 – PROBABILITY EXERCISE 13.3 Questions 1 to 14 Solutions

Question 1:

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?

Let's denote R as the event of drawing a red ball and B as the event of drawing a black ball.

Given that the first ball drawn is red, the probability of drawing a red ball again on the second draw is:

[ P(second ball is red | first ball is red) = P(R) × P(R | first ball is red) + P(B) × P(R | first ball is black) ]

[ = (1/2) × (6/12) + (1/2) × (5/12) = 11/24 ]
Question 2:

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.

Let A be the event that the ball is drawn from the first bag, and B be the event that the ball drawn is red.

Using Bayes' theorem:

[ P(A | B) = P(B | A) × P(A) / P(B) ]

[ = (4/8) × (1/2) / ((4/8 × 1/2) + (2/8 × 1/2)) = 4/6 = 2/3 ]
Question 3:

Of 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 examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostelier?

Let H be the event that a student resides in the hostel and A be the event that the student attains an A grade.

Using Bayes' theorem:

[ P(H | A) = P(A | H) × P(H) / P(A) ]

[ = (30/100) × (60/100) / ((30/100 × 60/100) + (20/100 × 40/100)) = 18/34 = 9/17 ]
Question 4:

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

Let K be the event that the student knows the answer and C be the event that the student answered it correctly.

Using Bayes' theorem:

[ P(K | C) = P(C | K) × P(K) / P(C) ]

[ = (3/4) × (1/2) / ((3/4 × 1/2) + (1/4 × 1/4)) = 6/7 ]

Question 5:

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 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?

Let D be the event that a person has the disease and T be the event that the test result is positive.

Using Bayes' theorem:

[ P(D | T) = P(T | D) × P(D) / P(T) ]

[ = 0.99 × 0.001 / ((0.001 × 0.99) + (0.999 × 0.005)) = 99/218 ]

Question 6:

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 time and the 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?

Let H_i denote the event that coin i is chosen (i = 1, 2, 3) and H denote the event of getting a head. We want to find P(H₁ | H).

Using Bayes' theorem:

P(H₁ | H) = (P(H | H₁) × P(H₁)) / P(H)

= (1 × 1/3) / [(1 × 1/3) + (3/4 × 1/3) + (1/2 × 1/3)] = 2/5

Question 7:

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident is 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?

Let A_i denote the event that the accident involves drivers of vehicle i (i = scooter, car, truck). We want to find P(A_scooter | accident).

Using Bayes' theorem:

P(A_scooter | accident) = (P(accident | A_scooter) × P(A_scooter)) / P(accident)

= (0.01 × 2000/12000) / [(0.01 × 2000/12000) + (0.03 × 4000/12000) + (0.15 × 6000/12000)]

= 1/7

Question 8:

A factory has two machines A and B. Past records show that machine A produced 60% of the items and machine B produced 40%. 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?

Let D denote the event that the chosen item is defective and M_i denote the event that the item is produced by machine i (i = A, B). We want to find P(M_B | D).

Using Bayes' theorem:

P(M_B | D) = (P(D | M_B) × P(M_B)) / P(D)

= (0.01 × 0.4) / [(0.02 × 0.6) + (0.01 × 0.4)] = 2/5

Question 9:

Two groups are competing for the position on the Board of directors of a corporation. 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.

Let G₁ and G₂ denote the events that the first and second groups win respectively, and P denote the event that a new product is introduced. We want to find P(P | G₂).

Using the law of total probability:

P(P) = P(G₁) × P(P | G₁) + P(G₂) × P(P | G₂)

= 0.6 × 0.7 + 0.4 × 0.3 = 0.6

Then using Bayes' theorem:

P(P | G₂) = ^{P(G₂ | P) × P(P)}⁄_P(G₂)

= ^{0.4 × 0.6}⁄_0.4 = 0.6

Question 10:

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?

Let D_i denote the event that the die shows number i (i = 1, 2, 3, 4) and H denote the event of getting exactly one head when tossing the coin. We want to find P(D₁ ∪ D₂ ∪ D₃ ∪ D₄ | H).

Using Bayes' theorem:

P(D₁ ∪ D₂ ∪ D₃ ∪ D₄ | H) = ^{P(H | D₁ ∪ D₂ ∪ D₃ ∪ D₄) × P(D₁ ∪ D₂ ∪ D₃ ∪ D₄)}⁄_P(H)

= ^{P(H | D₁ ∪ D₂ ∪ D₃ ∪ D₄) × (P(D₁) + P(D₂) + P(D₃) + P(D₄))}⁄_P(H)

= ^{∑_i=1⁴ P(D_i)}⁄_P(H) = ¹⁄₃

Question 11:

Part (a): The probability of obtaining an even prime number on each die when a pair of dice is rolled is 0 because there are no even prime numbers on a die.

Part (b): Two events A and B will be independent if P(A) * P(B) = P(A ∩ B). Therefore, the correct answer is (B) P(A|B) < P(A).

Question 12:

(a) The probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is ⁴⁄₅ since A's statement about a head appearing is most likely to be true.

(b)

(b) According to Bayes' theorem:

⁴/₅ × ¹/₂
----------------------------- = ⁴/₅
⁴/₅ × ¹/₂ + ¹/₂ × ¹/₅

Question 13:

(a) If A and B are two events such that A ⊂ B and P(B) ≠ 0, then the correct option is (A) P(B) P(A | B) / P(A).

(b) Since A is a subset of B, P(A | B) ≥ P(A). Therefore, the correct option is (C) P(A|B) ≥ P(A).

Question 14:

(a) The probability of obtaining an even prime number on each die when a pair of dice is rolled is 0. Since there are no even prime numbers on a standard six-sided die, the correct answer is (A) 0.

(b) Two events A and B will be independent if P(A ∩ B) = P(A) * P(B). Therefore, the correct option is (D) P(A) + P(B) = 1.