 # RBSE Maths Class 12 Chapter 16: Probability and Probability Distribution Important Questions and Solutions

RBSE Class 12 Maths Chapter 16 – Probability and Probability Distribution Important questions and solutions are available here. The important questions and solutions of this chapter given here cover all the important concepts. The RBSE Class 12 important questions and solutions given at BYJU’S will help the students of Class 12 in scoring maximum marks in their exams.

Chapter 16 of RBSE Class 12 contains five exercises, all of them involving various concepts of probability. In this chapter, different patterns of questions on finding the probability of events and probability distributions for the given situation are given. As probability deals with real-life situations, the chapter has different real-world scenarios for which the probability needs to be calculated.

### RBSE Maths Chapter 16: Exercise 16.1 Textbook Important Questions and Solutions

Question 1: If P(A) = 7/13, P(B) = 9/13 and P(A ⋂ B) = 4/13, then find P(A/B).

Solution:

Given,

P(A) = 7/13, P(B) = 9/13 and P(A ⋂ B) = 4/13

P(A/B) = P(A ⋂ B)/P(B)

= (4/13)/ (9/13)

= 4/9

Question 2: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, it is given that if at least one of them is a boy then find the probability that both the children will be a boy.

Solution:

Let A be the event that the family has at least one boy.

A = {BB, BG, GB}

i.e. n(A) = 3

And let B be the event that the family has two boys.

B = {BB}

i.e. n(B) = 1

Sample space = S {BB, BG, GB, GG}

n(S) = 4

Now, A ⋂ B = {BB}

n(A ⋂ B) = 1

Thus, P(A) = n(A)/n(S) = 3/4

P(B) = n(B)/n(S) = 1/4

P(A ⋂ B) = n(A ⋂ B)/n(S) = 1/4

Therefore,

P(B/A) = P(A ⋂ B)/P(A)

= (1/4)/ (3/4)

= 1/3

Hence, the required probability is 1/3.

### RBSE Maths Chapter 16: Exercise 16.2 Textbook Important Questions and Solutions

Question 3: If P(A) = 0.4, P(B) = p and P(A ⋃ B) = 0.6 and A, B are independent events, then find the value of p.

Solution:

Given,

P(A) = 0.4, P(B) = p and P(A ⋃ B) = 0.6

Also, A and B are independent events.

Therefore, P(A ∩ B) = P(A).P(B)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 0.6 = 0.4 + p – P(A)P(B)

0.6 = 0.4 + p – 0.4 × p

0.6 – 0.4 = (1 – 0.4)p

0.2 = 0.6 × p

⇒ p = 0.2/0.6

⇒ p = 1/3

Question 4: Three students are given the mathematical question to solve. Probability of solving the problem by the three are 1/2, 1/3 and 1/4. What is the probability that the question will be solved?

Solution:

We know that the problem will be solved if at least one of the three students can solve.

∴ Probability of solving the problem by one student = 1 – (1/2) = 1/2

Probability of not solving the problem by second student = 1 – (1/3) = 2/3

Similarly,

Probability of not solving the problem by third student = 1 – (1/4) = 3/4

Probability of not solving the problem by any one of them = (1/2) × (2/3) × (3/4) = 1/4

Hence, the probability of solving the problem by at least one of the three students = 1 – (1/4) = 3/4

### RBSE Maths Chapter 16: Exercise 16.3 Textbook Important Questions and Solutions

Question 5: 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.

Solution:

Let E1 be the event that the ball is drawn from the bag I and E2 be the event that the ball is drawn from the bag II.

A be the event that the drawn ball is red.

P(E1) = P(E2) = 1/2

Probability of getting a red ball from bag I = P(A/E1) = 3/7

Probability of getting a red ball from bag II = P(A/E2) = 5/11

Probability that the ball drawn from bag II is red = P(E2/A)

Using Bayes theorem,

P(E2/A) = [P(E2).P(A/E2)]/ [P(E1) P(A/E1) + P(E2) P(A/E2)]

= [(1/2) × (5/11)]/ [(1/2) × (3/7) + (1/2) × (5/11)]

= (5/12)/ [(3/14) + (5/22)]

= (5/22)/ (68/154)

= (5/22) × (154/68)

= 35/68

Question 6: A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3/10, 1/5, 1/10 and 2/5. The probabilities that he will be late are 1/4, 1/3 and 1/12 , if he comes by train, bus and scooter, respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

Solution:

Let A be the event that the doctor comes late to visit a patient.

And let E1, E2, E3 and E4 be the events that the Doctor comes by Train, Bus, Scooter and any-other means, respectively.

P(E1) = 3/10

P(E2) = 1/5

P(E3) = 1/10

P(E4) = 2/5

If the doctor comes by train, then probability of reaching late = P(A/E1) = 1/4

Similarly,

P(A/E2) = 1/3

P(A/E3) = 1/12

P(A/E4) = 0 (does not late by other means of transport)

Using Bayes theorem,

P(E1/A) = [P(E1).P(A/E1)]/ [P(E1) P(A/E1) + P(E2) P(A/E2) + P(E3) P(A/E3) + P(E4) P(A/E4)]

= [(3/10) × (1/4)]/ [(3/10) × (1/4) + (1/5) × (1/3) + (1/10) × (1/12) + (2/5) × 0]

= (3/40)/ [(3/40) + (1/15) + (1/120)]

= (3/40)/ (18/120)

= (3/40) × (120/18)

= 1/2

Hence, the required probability is 1/2.

### RBSE Maths Chapter 16: Exercise 16.4 Textbook Important Questions and Solutions

Question 7: Four rotten oranges by mistake are mixed with 16 good oranges. Two oranges are drawn and found to be rotten, find the probability distribution.

Solution:

Given,

4 bad oranges are mixed with 16 good oranges.

Thus, the total number of oranges = 4 + 16 = 20

Now, 2 bad oranges are to be chosen.

∴ Probability of choosing a bad orange

= 4/20 = 1/5

∴ Probability of choosing one good orange

= 1 – (1/5) = 4/5

Let X be the number of bad oranges.

P(X = 0) = P(GG)

P(X = 0) = (4/5) × (15/19) = 12/19

P(X = 1) = (4/20) × (16/19) + (16/20) × (4/1)

= (16/95) + (16/95)

= 32/95

P(X = 2) = P(BB)

= (4/20) × (3/19)

= 3/95

Hence, the probability distribution of bad oranges is:

 X 0 1 2 P(X) 12/19 32/95 3/95

### RBSE Maths Chapter 16: Exercise 16.5 Textbook Important Questions and Solutions

Question 8: An urn contains 5 white, 7 red and 8 black balls. If four balls are drawn with replacement then what is the probability that

(i) all balls are white

(ii) only three balls are white

(iii) none of the balls is white

(iv) at least three balls are white?

Solution:

Given,

Urn has 5 white, 7 red and 8 black balls.

(i) Total number of balls = 5 + 7 + 8 = 20

Number of white balls = 5

Probability of getting one white ball = 5/20 = 1/4

We know that all events are independent here.

∴ Probability of getting all white balls = (1/4) × (1/4) × (1/4) × (1/4) = 1/(4)4

(ii) Probability of drawing balls of which only 3 are white

= 3C1 × (1/4) × (1/4) × (1/4)

= 3 × 1/(4)3

(iii) Number of other balls = 7 + 8 = 15

∴ Probability of drawing one ball other than white = 15/20 = 3/4

P(none is white) = (3/4) × (3/4) × (3/4) × (3/4) = (3/4)4

(iv) P(at least 3 white balls) = P(3 white) + P(4 white)

= 3/(4)3 + (1/4)4

= (12 + 1)/(4)4

= 13/(4)4

### RBSE Maths Chapter 16: Additional Important Questions and Solutions

Question 1: The probability of a girl being a racer is 4 / 5. Find the probability of 4 girls being a racer out of 5 girls.

(A) (4/5)4 (1/5)

(B) 5C1 (1/5)(4/5)

(C) 5C4 (4/5)4 (1/5)

(D) None of these

Solution:

Probability that girl is a racer = 4/5

Thus, the probability that girls is not a racer

= 1 – (4/5) = 1/5

∴ Probability distribution of girls are racers

= [(4/5) + (1/5)]5

Hence, the probability that 4 girls are racers

= 5C4 (4/5)4 (1/5)

Question 2: A box contains 100 objects out of which 10 are defective. The probability of the given 5 objects , find the probability that none of them are defective is:

(A) (1/2)5

(B) 10-1

(C) 9/10

(D) (9/10)5

Solution:

Given,

Number of bulbs in box = 100

Number of defective bulbs = 10

Probability of defective bulbs = 10/100 = 1/10

Probability of non defective bulb = 1 – (1/10) = 9/10

Therefore, the probability that no bulb is defective out of 5 bulbs = P(X = 0)

= 5C0 (9/10)5 – 0 (1/10)0

= (9/10)5