NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 13 Probability

NCERT Solutions for Class 12 Maths Chapter 13 Miscellaneous Exercise – CBSE Free PDF Download

The Miscellaneous Exercise of NCERT Solutions for Class 12 Maths Chapter 13 – Probability is based on the following topics:

1. Introduction
2. Conditional Probability
3. Multiplication Theorem on Probability
4. Independent Events
5. Bayes’ Theorem
6. Random Variables and Its Probability Distributions
7. Bernoulli Trials and Binomial Distribution

Solving all the problems of this exercise will help the students practise different types of problems that cover all the concepts in the chapter.

NCERT Solution Class 12 Chapter 13 – Probability Miscellaneous Exercise

Download PDF

carouselExampleControls112

Previous





Next

Access Other Exercises of Class 12 Maths Chapter 13

Exercise 13.1 Solutions 17 Questions

Exercise 13.2 Solutions 18 Questions

Exercise 13.3 Solutions 14 Questions

Exercise 13.4 Solutions 17 Questions

Exercise 13.5 Solutions 15 Questions

Access Answers to NCERT Class 12 Maths Chapter 13 Miscellaneous

1. A and B are two events such that P (A) ≠ 0. Find P (B|A), if:
(i) A is a subset of B
(ii) A ∩ B = φ

Solution:

2. A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Solution:

(i) According to the question, if the couple has two children, then the sample space is

S = {(b, b), (b, g), (g, b), (g, g)}

Assume that A denote the event of both children having male and B denote the event of having at least one of the male children,

Thus, we have

(ii) Assume that C denote the event of having both children females and D denotes the event of having an elder child as female.

∴ C = {(g, g)}

P (C) = ¼

And, D = {(g, b), (g, g)}

P (D) = (2/4)

3. Suppose that 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.

Solution:

Given that, 5% of men and 0.25% of women have grey hair.

∴ Total % of people having grey hair = 5 + 0.25

= 5.25 %

Hence, the probability of having a selected person male having grey hair, P = 5/25 = 20/21

4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution:

Given that, 90% of the people are right-handed.

Let p denotes the probability of people that are right-handed and q denotes the probability of people that are left-handed.

p = 9/10 and q = 1 – 9/10 = 1/10

Now, by using the binomial distribution probability of having more than 6 right-handed people can be given as

5. An urn contains 25 balls, of which 10 balls bear the mark ‘X’ and the remaining 15 bear the mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down, and it is replaced. If 6 balls are drawn in this way, find the probability that:
(i) All will bear the ‘X’ mark.
(ii) Not more than 2 will bear the ‘Y’ mark.
(iii) At least one ball will bear the ‘Y’ mark.
(iv) The number of balls with the ‘X’ mark and the ‘Y’ mark will be equal.

Solution:

(i) It is given in the question that

Total number of balls in the urn = 25

Number of balls bearing mark ‘X’ = 10

Number of balls bearing mark ‘Y’ = 15

Let p denotes the probability of balls bearing the mark ‘X’ and q denotes the probability of balls bearing the mark ‘Y’.

p = 10/25 = 2/5 and q = 15/25 = 3/5

Now, 6 balls are drawn with replacement. Hence, the number of trials are Bernoulli triangle.

Assume Z be the random variable that represents the number of balls bearing the ‘Y’ mark in the trials.

∴ Z has a binomial distribution where n = 6 and p = 2/5

6. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles?

Solution:

Assume that p is the probability that the player will clear the hurdle while q is the probability that the player will knock down the hurdle.

∴ p = 5/6 and q = 1 – 5/6 = 1/6

Let us also assume X be the random variable that represents the number of times the player will knock down the hurdle.

7. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Solution:

From the given question, it is clear that

Probability of getting a six in a throw of die = 1/6

And, the probability of not getting a six = 5/6

Let us assume, p = 1/6 and q = 5/6

Now, we have

The probability that the 2 sixes come in the first five throws of the die

8. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Solution:

We know that in a leap year, there are a total of 366 days, 52 weeks and 2 days.

Now, in 52 weeks, there is a total of 52 Tuesdays.

∴ The probability that the leap year will contain 53 Tuesdays is equal to the probability of remaining 2 days will be Tuesdays.

Thus, the remaining two days can be

(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and Saturday), (Saturday and Sunday) and (Sunday and Monday)

∴ Total number of cases = 7

Cases in which Tuesday can come = 2

Hence, the probability (leap year having 53 Tuesdays) = 2/7

9. An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

Solution:

Given that probability of failure = x

And, the probability of success = 2x

∴ x + 2x = 1

3x = 1

X = 1/3

2x = 2/3

Assume p = 1/3 and q = 2/3

Also, X be the random variable that represents the number of trials.

Hence, by binomial distribution, we have

10. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Solution:

Let us assume that man tosses the coin n times. Thus, n tosses are the Bernoulli trials.

∴ Probability of getting head at the toss of the coin = ½

Let us assume, p = ½ and q = ½

11. In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but quit as and when he got a six. Find the expected value of the amount he wins/loses.

Solution:

For the situation given in the equation, we have

Probability of getting a six in a throw of a die = 1/6

Also, the probability of not getting a 6 = 5/6

Now, there are three cases from which the expected value of the amount he wins can be calculated.

(i) the First case is that if he gets a six on his first through, then the required probability will be 1/6.

∴ Amount received by him = Rs. 1

(ii) Secondly, if he gets six on his second throw, then the probability = (5/6 × 1/6)

= 5/36

∴ Amount received by him = – Rs. 1 + Rs. 1

= 0

(iii) Lastly, if he does not get six in his first two throws and gets six in his third throw, then the probability  = 5/6 × 5/6 × 1/6

= 25/216

∴ Amount received by him = – Rs. 1 – Rs. 1 + Rs. 1

= – 1

Hence, the expected value that he can win = 1/6 – 25/216

= (36 – 25)/216

= 11/216

12. Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

One of the boxes has been selected at random, and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A, box B and box C?

Solution:

Let us assume R be the event of drawing red marbles.

Let us also assume E_A, E_B and E_C denote the boxes A, B and C, respectively.

Given that,

Total number of marbles = 40

Also, the total number of red marbles = 15

P (R) = 15/40

= 3/8

Probability of taking out the red marble from box A,

13. Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30%, and a prescription for a certain drug reduces its chances by 25%. At a time, a patient can choose any one 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.

Solution:

Let us assume X denotes the events having a person’s heart attack.

A₁ denote events having the selected person follow the course of yoga and meditation.

And A₂ denote the events having the person adopted the drug prescription.

It is given in the question that

P (X) = 0.40

14. If each element of a second-order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2.)

Solution:

From the question, we have

Total number of determinants of second order where the element being or 1 = (2)⁴

= 16

Now, we have the value of determinants is positive in the following cases:

∴ Required probability = 3/16

15. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P (A fails) = 0.2
P (B fails alone) = 0.15
P (A and B fail) = 0.15
Evaluate the following probabilities:
(i) P (A fails | B has failed)
(ii) P (A fails alone)

Solution:

(i) Let us assume the event which is failed by A is denoted by E_A

And, the event which is failed by B is denoted by E_B

It is given in the question that,

The event failed by A, P (E_A) = 0.2

(ii) We have a probability where A fails alone

= 0.2 – 0.15

= 0.05

16. 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 a 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.
Choose the correct answer in each of the following:

Solution:

Let us first assume A₁ denote the events that a red ball is transferred from the bag I to II.

And A₂ denote the event that a black ball is transferred from the bag I to II.

∴ P (A₁) = 3/7

And, P (A₂) = 4/7

Let X be the event that the drawn ball is red.

∴ when the red ball is transferred from the bag I to II,

17. If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
A. A ⊂ B
B. B ⊂ A
C. B = φ
D. A = φ

Solution:

A. A ⊂ B

Explanation:

18. If P (A|B) > P (A), then which of the following is correct:
A. P (B|A) < P (B)
B. P (A ∩ B) < P (A) . P (B)
C. P (B|A) > P (B)
D. P (B|A) = P (B)

Solution:

C. P (B|A) > P (B)

Explanation:

19. If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
A. P(B|A) = 1
B. P(A|B) = 1
C. P(B|A) = 0
D. P(A|B) = 0

Solution:

B. P (A|B) = 1
Explanation:

Also, explore –

NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

NCERT Solutions