NCERT Solutions for Class 12 Maths Chapter 13 Probability: You have already studied the basics of probability in our previous classes like probability as a measure of uncertainty of events in a random experiment, addition rule of probability etc. NCERT solutions for class 12 maths chapter 13 probability will build your base to study probability theory in higher studies. It includes concepts of discrete probability, computational probability and stochastic process. The important topics like conditional probability, Bayes' theorem, multiplication rule of probability, and independence of events are covered in this chapter. Questions from all these topics are covered in CBSE NCERT solutions for class 12 maths chapter 13 probability. Also, you will learn some important concepts of the random variable, probability distribution, and the mean and variance of a probability distribution in this chapter. Solutions of NCERT for class 12 maths chapter 13 probability will help you to learn the concept of probability distribution which will be required in higher study. Check all NCERT solutions from class 6 to 12 at a single place, which will help you get a better understanding of concepts in a much easy way.
Generally, two questions( 8 marks) are asked from this chapter in 12th board final examination. You can score this 8 marks very easily with the help of NCERT solutions for class 12 maths chapter 13 probability. In NCERT textbook there are 37 solved examples are given, so you can understand the concept easily. In this chapter, there is a total of 81 questions in 5 exercises. You should try to solve every question on your own. If you are not able to do, you can take help of these CBSE NCERT solutions for class 12 maths chapter 13 probability. These NCERT questions are solved and explained in a stepbystep manner, so it can be understood very easily. This chapter requires lots of practice to understand it better. So, you are advised to solve miscellaneous exercise too.
The conditional probability of an event E, given the occurrence of the event F is given by

13.1 Introduction
13.2 Conditional Probability
13.2.1 Properties of conditional probability
13.3 Multiplication Theorem on Probability
13.4 Independent Events
13.5 Bayes' Theorem
13.5.1 Partition of a sample space
3.5.2 Theorem of total probability
13.6 Random Variables and its Probability Distributions
13.6.1 Probability distribution of a random variable
13.6.2 Mean of a random variable
13.6.3 Variance of a random variable
13.7 Bernoulli Trials and Binomial Distribution
13.7.1 Bernoulli trials
13.7.2 Binomial distribution
Answer:
It is given that and
Question:6 A coin is tossed three times, where
(i)E : head on third toss ,F : heads on first two tosses
Answer:
The sample space S when a coin is tossed three times is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space (S) has 8 elements.
Total number of outcomes
According to question
E: head on third toss, F: heads on first two tosses
Question:6 A coin is tossed three times, where
(ii)E : at least two heads ,F : at most two heads
Answer:
The sample space S when a coin is tossed three times is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space (S) has 8 elements.
Total number of outcomes
According to question
E : at least two heads , F : at most two heads
Question:6 A coin is tossed three times, where
(iii)E : at most two tails ,F : at least one tail
Answer:
The sample space S when a coin is tossed three times is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space (S) has 8 elements.
Total number of outcomes
According to question
E: at most two tails, F: at least one tail
Question:7 Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
Answer:
E : tail appears on one coin, F : one coin shows head
Total outcomes =4
Question:7 Two coins are tossed once, where
(ii)E : no tail appears,F : no head appears
Answer:
E : no tail appears, F : no head appears
Total outcomes =4
Question:8 A die is thrown three times,
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses
Answer:
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses
Total outcomes
Question:9 Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle
Answer:
E : son on one end, F : father in middle
Total outcomes
Let S be son, M be mother and F be father.
Then we have,
Question:10 A black and a red dice are rolled.
Answer:
A black and a red dice are rolled.
Total outcomes
Let the A be event obtaining a sum greater than and B be a event that the black die resulted in a
Question:10 A black and a red dice are rolled.
Answer:
A black and a red dice are rolled.
Total outcomes
Let the A be event obtaining a sum 8 and B be a event thatthat the red die resulted in a number less than .
Red dice is rolled after black dice.
Question:11 A fair die is rolled. Consider events and Find
Answer:
A fair die is rolled.
Total oucomes
and
Question:11 A fair die is rolled. Consider events and Find
Answer:
A fair die is rolled.
Total oucomes
,
Question:11 A fair die is rolled. Consider events and Find
Answer:
A fair die is rolled.
Total oucomes
and
,
Answer:
Assume that each born child is equally likely to be a boy or a girl.
Let first and second girl are denoted by respectively also first and second boy are denoted by
If a family has two children, then total outcomes
Let A= both are girls
and B= the youngest is a girl =
Therefore, the required probability is 1/2
Answer:
Assume that each born child is equally likely to be a boy or a girl.
Let first and second girl are denoted by respectively also first and second boy are denoted by
If a family has two children, then total outcomes
Let A= both are girls
and C= at least one is a girl =
Answer:
An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.
Total number of questions
Let A = question be easy.
Let B = multiple choice question
easy multiple questions
Therefore, the required probability is 5/9
Answer:
Two dice are thrown.
Total outcomes
Let A be the event ‘the sum of numbers on the dice is 4.
Let B be the event that two numbers appearing on throwing two dice are different.
Therefore, the required probability is 1/15
Answer:
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin.
Total outcomes
Total number of outcomes =20
Let A be a event when coin shows a tail.
Let B be a event that ‘at least one die shows a 3’.
Question:16 In the following Exercise 16 choose the correct answer:
Answer:
It is given that
Hence, is not defined .
Thus, correct option is C.
Question:17 In the following Exercise 17 choose the correct answer:
If and are events such that then
Answer:
It is given that
Hence, option D is correct.
Question:1 If and find if and are independent events.
Answer:
and
Given : and are independent events.
So we have,
Answer:
Two cards are drawn at random and without replacement from a pack of 52 playing cards.
There are 26 black cards in a pack of 52.
Let be the probability that first cards is black.
Then, we have
Let be the probability that second cards is black.
Then, we have
The probability that both the cards are black
Answer:
Total oranges = 15
Good oranges = 12
Bad oranges = 3
Let be the probability that first orange is good.
The, we have
Let be the probability that second orange is good.
Let be the probability that third orange is good.
The probability that a box will be approved for sale
Answer:
A fair coin and an unbiased die are tossed,then total outputs are:
A is the event ‘head appears on the coin’ .
Total outcomes of A are :
B is the event ‘3 on the die’.
Total outcomes of B are :
Also,
Hence, A and B are independent events.
Answer:
Total outcomes .
is the event, ‘the number is even,’
Outcomes of A
is the event, ‘the number is red’.
Outcomes of B
Also,
Thus, both the events A and B are not independent.
Question:6 Let and be events with and Are E and F independent?
Answer:
Given :
and
For events E and F to be independent , we need
Hence, E and F are not indepent events.
Question:7 Given that the events and are such that and Find if they are
Answer:
Given,
Also, A and B are mutually exclusive means .
Question:7 Given that the events and are such that and Find p if they are
Answer:
Given,
Also, A and B are independent events means
. Also
Question:8 Let A and B be independent events with and Find
Answer:
and
Given : A and B be independent events
So, we have
Question:8 Let and be independent events with and Find
Answer:
and
Given : A and B be independent events
So, we have
We have,
Question:8 Let and be independent events with and Find
Answer:
and
Given : A and B be independent events
So, we have
Question:8 Let A and B be independent events with and Find
Answer:
and
Given : A and B be independent events
So, we have
Question:9 If and are two events such that and find
Answer:
If and are two events such that and
use,
Question:10 Events A and B are such that and State whether and are independent ?
Answer:
If and are two events such that and
As we can see
Hence, A and B are not independent.
Question:11 Given two independent events and such that Find
Answer:
Given two independent events and .
Also , we know
Question:11 Given two independent events A and B such that Find
Answer:
Given two independent events and .
Question:12 A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer:
A die is tossed thrice.
Outcomes
Odd numbers
The probability of getting an odd number at first throw
The probability of getting an even number
Probability of getting even number three times
The probability of getting an odd number at least once = 1  the probability of getting an odd number in none of throw
= 1  probability of getting even number three times
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
The probability of getting a red ball in first draw
The ball is repleced after drawing first ball.
The probability of getting a red ball in second draw
the probability that both balls are red
(ii) first ball is black and second is red.
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
The probability of getting a black ball in the first draw
The ball is replaced after drawing the first ball.
The probability of getting a red ball in the second draw
the probability that the first ball is black and the second is red
(iii) one of them is black and other is red.
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
Let the first ball is black and the second ball is red.
The probability of getting a black ball in the first draw
The ball is replaced after drawing the first ball.
The probability of getting a red ball in the second draw
the probability that the first ball is black and the second is red
Let the first ball is red and the second ball is black.
The probability of getting a red ball in the first draw
The probability of getting a black ball in the second draw
the probability that the first ball is red and the second is black
Thus,
The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probability that the first ball is red and the second is black
Answer:
and
Since, problem is solved independently by A and B,
probability that the problem is solved
(ii) exactly one of them solves the problem
Answer:
and
,
,
probability that exactly one of them solves the problem
probability that exactly one of them solves the problem
(i) E : ‘the card drawn is a spade’
F : ‘the card drawn is an ace’
Answer:
One card is drawn at random from a well shuffled deck of cards
Total ace = 4
total spades =13
E : ‘the card drawn is a spade
F : ‘the card drawn is an ace’
a card which is spade and ace = 1
Hence, E and F are indepentdent events .
F : ‘the card drawn is a king’
Answer:
One card is drawn at random from a well shuffled deck of cards
Total black card = 26
total king =4
E : ‘the card drawn is black’
F : ‘the card drawn is a king’
a card which is black and king = 2
Hence, E and F are indepentdent events .
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
Answer:
One card is drawn at random from a well shuffled deck of cards
Total king or queen = 8
total queen or jack = 8
E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
a card which is queen = 4
Hence, E and F are not indepentdent events
(a) Find the probability that she reads neither Hindi nor English newspapers
Answer:
H : of the students read Hindi newspaper,
E : read English newspaper and
read both Hindi and English newspapers.
the probability that she reads neither Hindi nor English newspapers
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
Answer:
H : of the students read Hindi newspaper,
E : read English newspaper and
read both Hindi and English newspapers.
The probability that she reads English newspape if she reads Hindi newspaper
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer:
H : of the students read Hindi newspaper,
E : read English newspaper and
read both Hindi and English newspapers.
the probability that she reads Hindi newspaper if she reads English newspaper
Question:17 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
Answer:
when a pair of dice is rolled, total outcomes
Even prime number
The probability of obtaining an even prime number on each die
Option D is correct.
Question:18 Two events A and B will be independent, if
(A) and are mutually exclusive
Answer:
Two events A and B will be independent, if
Or
Option B is correct.
Answer:
Black balls = 5
Red balls = 5
Total balls = 10
CASE 1 Let red ball be drawn in first attempt.
Now two red balls are added in urn .
Now red balls = 7, black balls = 5
Total balls = 12
CASE 2
Let black ball be drawn in first attempt.
Now two black balls are added in urn .
Now red balls = 5, black balls = 7
Total balls = 12
the probability that the second ball is red =
Answer:
BAG 1 : Red balls =4 Black balls=4 Total balls = 8
BAG 2 : Red balls = 2 Black balls = 6 Total balls = 8
B1 : selecting bag 1
B2 : selecting bag 2
Let R be a event of getting red ball
probability that the ball is drawn from the first bag,
given that it is red is .
Using Baye's theorem, we have
Answer:
H : reside in hostel
D : day scholars
A : students who attain grade A
By Bayes theorem :
Answer:
A : Student knows answer.
B : Student guess the answer
C : Answer is correct
By Bayes theorem :
Answer:
A : Person selected is having the disease
B : Person selected is not having the disease.
C :Blood result is positive.
By Bayes theorem :
Answer:
Given : A : chossing a two headed coin
B : chossing a biased coin
C : chossing a unbiased coin
D : event that coin tossed show head.
Biased coin that comes up heads of the time.
Answer:
Let A : scooter drivers = 2000
B : car drivers = 4000
C : truck drivers = 6000
Total drivers = 12000
D : the event that person meets with an accident.
Answer:
A : Items produced by machine A
B : Items produced by machine B
X : Produced item found to be defective.
Hence, the probability that defective item was produced by machine =
.
Answer:
A: the first groups will win
B: the second groups will win
X: Event of introducing a new product.
Probability of introducing a new product if the first group wins :
Probability of introducing a new product if the second group wins :
Hence, the probability that the new product introduced was by the second group :
Answer:
Let, A: Outcome on die is 5 or 6.
B: Outcome on die is 1,2,3,4
X: Event of getting exactly one head.
Probability of getting exactly one head when she tosses a coin three times :
Probability of getting exactly one head when she tosses a coin one time :
Hence, the probability that she threw or with the die =
Answer:
Let A: time consumed by machine A
B: time consumed by machine B
C: time consumed by machine C
Total drivers = 12000
D: Event of producing defective items
Hence, the probability that defective item was produced by =
Answer:
Let A : Event of choosing a diamond card.
B : Event of not choosing a diamond card.
X : The lost card.
If lost card is diamond then 12 diamond cards are left out of 51 cards.
Two diamond cards are drawn out of 12 diamond cards in ways.
Similarly, two cards are drawn out of 51 cards in ways.
Probablity of getting two diamond cards when one diamond is lost :
If lost card is not diamond then 13 diamond cards are left out of 51 cards.
Two diamond cards are drawn out of 13 diamond cards in ways.
Similarly, two cards are drawn out of 51 cards in ways.
Probablity of getting two diamond cards when one diamond is not lost :
The probability of the lost card being a diamond :
Hence, the probability of the lost card being a diamond :
Answer:
Let A : A speaks truth
B : A speaks false
X : Event that head appears.
A coin is tossed , outcomes are head or tail.
Probability of getting head whether A speaks thruth or not is
The probability that actually there was head is
Hence, option A is correct.
Question:14 If and are two events such that and then which of the following is correct?
Answer:
If and then
Also,
We know that
Hence, we can see option C is correct.
Question:1(i) State which the following are not the probability distributions of a random variable. Give reasons for your answer.
Answer:
As we know the sum of probabilities of a probability distribution is 1.
Sum of probabilities
The given table is the probability distributions of a random variable.
Question:1(ii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Answer:
As we know probabilities cannot be negative for a probability distribution .
The given table is not a the probability distributions of a random variable.
Question:1(iii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Answer:
As we know sum of probabilities of a probability distribution is 1.
Sum of probablities
The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.
Question:1(iv) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Answer:
As we know sum of probabilities of a probability distribution is 1.
Sum of probablities
The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.
Answer:
B = black balls
R = red balls
The two balls can be selected as BR,BB,RB,RR.
X = number of black balls.
Hence, possible values of X can be 0, 1 and 2.
Yes, X is a random variable.
Answer:
The difference between the number of heads and the number of tails obtained when a coin is tossed times are :
Thus, possible values of X are 0, 2, 4 and 6.
Question:4(i) Find the probability distribution of
number of heads in two tosses of a coin.
Answer:
When coin is tossed twice then sample space
Let X be number of heads.
X can take values of 0,1,2.
Table is as shown :
X 
0 
1 
2 
P(X) 



Question:4(ii) Find the probability distribution of
number of tails in the simultaneous tosses of three coins.
Answer:
When 3 coins are simultaneous tossed then sample space
Let X be number of tails.
X can be 0,1,2,3
X can take values of 0,1,2.
Table is as shown :
X 
0 
1 
2 
3 
P(X) 




Question:4(iii) Find the probability distribution of
number of heads in four tosses of a coin.
Answer:
When coin is tossed 4 times then sample space
Let X be number of heads.
X can be 0,1,2,3,4
Table is as shown :
X 
0 
1 
2 
3 
4 
P(X) 





Question:5(i) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
Answer:
When a die is tossed twice , total outcomes = 36
Number less than or equal to 4 in both toss :
Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss:
Number less than 4 in both tosses :
Probability distribution is as :
X 
0 
1 
2 
P(X) 



Question:5(ii) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
six appears on at least one die.
Answer:
When a die is tossed twice , total outcomes = 36
Six does not appear on any of the die :
Six appear on atleast one die :
Probability distribution is as :
X 
0 
1 
P(X) 


Answer:
Total bulbs = 30
defective bulbs = 6
Non defective bulbs
bulbs is drawn at random with replacement.
Let X : number of defective bulbs
4 Non defective bulbs and 0 defective bulbs :
3 Non defective bulbs and 1 defective bulbs :
2 Non defective bulbs and 2 defective bulbs :
1 Non defective bulbs and 3 defective bulbs :
0 Non defective bulbs and 4 defective bulbs :
the probability distribution of the number of defective bulbs is as :
X 
0 
1 
2 
3 
4 
P(X) 





Answer:
the coin is tossed twice, total outcomes =4
probability of getting a tail be x.
i.e.
Then
and
Let X : number of tails
No tail :
1 tail :
2 tail :
the probability distribution of number of tails are
X 
0 
1 
2 