Pearson | PTE
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
NCERT Solutions for Class 12 Maths Chapter 13 Probability are provided here. You have already studied the basics of probability class 12 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 therefore probability class 12 ncert solutions become very important.
CBSE Class 10 date sheet 2025 for upcoming exams.
Exam Dates | Subjects |
February 28, 2025 | Hindi Course A, Hindi Course B |
March 1, 2025 | Painting |
March 3, 2025 | Healthcare |
March 5, 2025 | Elements of Business, Retail |
March 6, 2025 | Tibetian, German, National Cadet Corps, Telugu-Telangana, Bodo, Tangkhul, Japanese, Bhoti, Spanish, Bhutia Kashmiri, Mizo, Bahasa Melayu |
March 10, 2025 | Mathematics Standard, Mathematics Basic |
March 12, 2025 | Telegu, Arabic, Russian, Persian, Lepcha, Nepali, Limboo, German, Carnatic Music (Melodic Instruments), Carnatic Music (Vocal), Carnatic Music (Percussion Instruments), Thai, Hindustani Music (Melodic Instruments), Hindustani Music (Vocal) |
March 13, 2025 | Home Science |
March 17, 2025 | Punjabi, Sindhi, Malayalam, Odia, Assamese, Kannada, Kokborok |
March 18, 2025 | Computer Applications, Information and Technology, Artificial Intelligence |
NCERT Class 12 Maths Chapter 13 solutions include 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 of NCERT Class 12 maths probability books. Questions from all these topics are covered in 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 13 NCERT Class 12 maths solutions PDF. class 12 maths ncert solutions pdf will help you to learn the concept of probability distribution which will be required in higher study. NCERT solutions help students to understan the concepts in a much easy way. Here you will get NCERT solutions for class 12 other subjects also.
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
Also read:
>> Conditional Probability: Conditional probability is the likelihood of an event occurring based on the occurrence of a preceding event. For two events A and B with the same sample space, the conditional probability of event A given that B has occurred (P(A|B)) is defined as:
P(A|B) = P(A ∩ B) / P(B) (when P(B) ≠ 0)
Other conditional probability relationships:
P(S|F) = P(F|F) = 1
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
P(E'|F) = 1 − P(E|F)
>> Multiplication Rule: The multiplication rule relates the probability of two events E and F in a sample space S:
P(E ∩ F) = P(E) P(F|E) = P(F) P(E|F) (when P(E) ≠ 0 and P(F) ≠ 0)
>> Independent Events: Two experiments are considered independent if the probability of the events E and F occurring simultaneously is the product of their individual probabilities:
P(E ∩ F) = P(E) * P(F)
>> Bayes’ Theorem: Bayes’ theorem deals with events E1, E2, …, En that form a partition of the sample space S. It allows the calculation of the probability of event Ei given event A:
P(Ei|A) = [P(Ei) * P(A|Ei)] / ∑[P(Ej) * P(A|Ej)], for i = 1, 2, …, n
>> Theorem of Total Probability: Given a partition E1, E2, …, En of the sample space and an event A, the theorem of total probability states:
P(A) = P(E1) * P(A|E1) + P(E2) * P(A|E2) + … + P(En) * P(A|En)
>> Random Variables and their Probability Distributions: A random variable is a real-valued function whose domain is a sample space. The probability distribution of a random variable X consists of possible values x1, x2, …, xn and their corresponding probabilities p1, p2, …, pn:
E(X) = μ = ∑(xi * pi)
Var(X) = σ² = ∑((xi - μ)² * pi)
σ = √Var(X)
Free download NCERT Class 12 Maths Chapter 13 Question Answer Probability for CBSE Exam.
NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.1
Answer:
It is given that
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
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
Answer:
A fair die is rolled.
Total oucomes
Question:11 A fair die is rolled. Consider events
Answer:
A fair die is rolled.
Total oucomes
Question:11 A fair die is rolled. Consider events
Answer:
A fair die is rolled.
Total oucomes
Answer:
Assume that each born child is equally likely to be a boy or a girl.
Let first and second girl 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
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
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
Answer:
It is given that
Hence, option D is correct.
NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.2
Question:1 If
Answer:
Given :
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
Then, we have
Let
Then, we have
The probability that both the cards are black
Answer:
Total oranges = 15
Good oranges = 12
Bad oranges = 3
Let
The, we have
Let
Let
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
Outcomes of A
Outcomes of B
Also,
Thus, both the events A and B are not independent.
Question:6 Let
Answer:
Given :
For events E and F to be independent , we need
Hence, E and F are not indepent events.
Question:7 Given that the events
Answer:
Given,
Also, A and B are mutually exclusive means
Question:7 Given that the events
Answer:
Given,
Also, A and B are independent events means
Question:8 Let A and B be independent events with
Answer:
Given : A and B be independent events
So, we have
Question:8 Let
Answer:
Given : A and B be independent events
So, we have
We have,
Question:8 Let
Answer:
Given : A and B be independent events
So, we have
Question:8 Let A and B be independent events with
Answer:
Given : A and B be independent events
So, we have
Question:9 If
Answer:
If
Question:10 Events A and B are such that
Answer:
If
As we can see
Hence, A and B are not independent.
Question:11 Given two independent events
Answer:
Given two independent events
Also , we know
Question:11 Given two independent events A and B such that
Answer:
Given two independent events
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:
Since, problem is solved independently by A and B,
probability that the problem is solved
(ii) exactly one of them solves the problem
Answer:
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
Total ace = 4
total spades =13
E : ‘the card drawn is a spade
F : ‘the card drawn is an ace’
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
Total black card = 26
total king =4
E : ‘the card drawn is black’
F : ‘the card drawn is a king’
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
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’.
Hence, E and F are not indepentdent events
(a) Find the probability that she reads neither Hindi nor English newspapers
Answer:
H :
E :
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 :
E :
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 :
E :
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)
Answer:
Two events A and B will be independent, if
Or
Option B is correct.
NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.3
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
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
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
Similarly, two cards are drawn out of 51 cards in
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
Similarly, two cards are drawn out of 51 cards in
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
Answer:
If
Also,
We know that
Hence, we can see option C is correct.
NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.4
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
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 .
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
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
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
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 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.
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
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
Let X : number of tails
No tail :
1 tail :
2 tail :
the probability distribution of number of tails are
X | 0 | 1 | 2 |
P(X) | | | |
Question:8(i) A random variable X has the following probability distribution:
Answer:
Sum of probabilities of probability distribution of random variable is 1.
Question:9(a) The random variable X has a probability distribution P(X) of the following form, where k is some number :
Answer:
Sum of probabilities of probability distribution of random variable is 1.
Question:10 Find the mean number of heads in three tosses of a fair coin.
Answer:
Let X be the success of getting head.
When 3 coins are tossed then sample space
The probability distribution is as
X | 0 | 1 | 2 | 3 |
P(X) | | | | |
mean number of heads :
Question:11 Two dice are thrown simultaneously. If
Answer:
X can be 0,1,2.
Getting six on one time when thrown twice :
Getting six on both dice :
X | 0 | 1 | 2 |
P(X) | | | |
Expectation of X = E(X)
Answer:
Two numbers are selected at random (without replacement) from the first six positive integers in
X can be 2,3,4,5,6.
X=2, obsevations :
X=3, obsevations :
X=4, obsevations :
X=5, obsevations :
X=6, obsevations :
Probability distribution is as follows:
X | 2 | 3 | 4 | 5 | 6 |
P(X) | | | | | |
Answer:
Total observations = 36
X can be 2,3,4,5,6,7,8,9,10,11,12
Probability distribution is as follows :
X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
P(X) | | | | | | | | | | | |
Standard deviation =
Answer:
Total students = 15
probability of selecting a student :
The information given can be represented as frequency table :
X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |
Probability distribution is as :
X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
P(X) | | | | | | | | |
Answer:
Given :
Probability distribution is as :
X | 0 | 1 |
P(X) | 0.3 | 0.7 |
Answer:
X is number representing on die.
Total observations = 6
X | 1 | 2 | 5 |
P(X) | | | |
Option B is correct.
Answer:
X be number od aces obtained.
X can be 0,1,2
There 52 cards and 4 aces, 48 are non-ace cards.
The probability distribution is as :
X | 0 | 1 | 2 |
P(X) | | | |
Option D is correct.
NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.5
Question:1(i) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
Answer:
X be the number of success of getting an odd number.
X has a binomial distribution.
Question:1(ii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
Answer:
X be a number of success of getting an odd number.
X has a binomial distribution.
Question:1(iii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
Answer:
X be a number of success of getting an odd number.
X has a binomial distribution.
Answer:
A pair of dice is thrown
Probability of getting doublet in a throw of pair of dice :
X has a binomial distribution,n=4
Put x = 2
Answer:
There are
X denotes the number of defective items in a sample of 10.
X has a binomial distribution, n=10.
Question:4(i) Five cards are drawn successively with replacement from a well-shuffled deck of
all the five cards are spades?
Answer:
Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of
We have 13 spades.
X has a binomial distribution,n=5.
Put X=5 ,
Question:4(ii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
Answer:
Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of
We have 13 spades.
X has a binomial distribution,n=5.
Put X=3 ,
Question:4(iii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
Answer:
Let X represent number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of
We have 13 spades .
X has a binomial distribution,n=5.
Put X=0 ,
Question:5(i) The probability that a bulb produced by a factory will fuse after
none will fuse after
Answer:
Let X represent number of bulb that will fuse after
X has a binomial distribution,n=5.
Put X=0 ,
Question:5(ii) The probability that a bulb produced by a factory will fuse after
not more than one will fuse after
Answer:
Let X represent a number of the bulb that will fuse after
X has a binomial distribution,n=5.
Put
Question:5(iii) The probability that a bulb produced by a factory will fuse after
more than one will fuse after
Answer:
Let X represent number of bulb that will fuse after
X has a binomial distribution,n=5.
Put
Question:5(iv) The probability that a bulb produced by a factory will fuse after
at least one will fuse after
Answer:
Let X represent number of bulb that will fuse after
X has a binomial distribution,n=5.
Put
Answer:
Let X denote a number of balls marked with digit 0 among 4 balls drawn.
Balls are drawn with replacement.
X has a binomial distribution,n=4.
Put X = 0,
Answer:
Let X represent the number of correctly answered questions out of 20 questions.
The coin falls heads, he answers 'true'; if it falls tails, he answers 'false'.
X has a binomial distribution,n=20
Question:8 Suppose X has a binomial distribution
(Hint :
Answer:
X is a random variable whose binomial distribution is
Here , n=6 and
Answer:
Let X represent number of correct answers by guessing in set of 5 multiple choice questions.
Probability of getting a correct answer :
X has a binomial distribution,n=5.
Answer:
Let X represent number of winning prizes in 50 lotteries .
X has a binomial distribution,n=50.
Answer:
Let X represent number of winning prizes in 50 lotteries .
X has a binomial distribution,n=50.
Answer:
Let X represent number of winning prizes in 50 lotteries.
X has a binomial distribution,n=50.
Question:11 Find the probability of getting
Answer:
Let X represent number of times getting 5 in 7 throws of a die.
Probability of getting 5 in single throw of die=P
X has a binomial distribution,n=7
Question:12 Find the probability of throwing at most
Answer:
Let X represent number of times getting 2 six in 6 throws of a die.
Probability of getting 6 in single throw of die=P
X has a binomial distribution,n=6
Answer:
Let X represent a number of times selecting defective items out of 12 articles.
Probability of getting a defective item =P
X has a binomial distribution,n=12
Answer:
Let X represent a number of defective bulbs out of 5 bulbs.
Probability of getting a defective bulb =P
X has a binomial distribution,n=5
The correct answer is C.
In the following, choose the correct answer:
Answer:
Let X represent number students out of 5 who are swimmers.
Probability of student who are not swimmers =q
X has a binomial distribution,n=5
Option A is correct.
NCERT solutions for class 12 maths chapter 13 probability-Miscellaneous Exercise
Question:1(i) A and B are two events such that
Answer:
A and B are two events such that
Question:2(i) A couple has two children,
Answer:
A couple has two children,
sample space
Let A be both children are males and B is at least one of the children is male.
Question:2(ii) A couple has two children,
Answer:
A couple has two children,
sample space
Let A be both children are females and B be the elder child is a female.
Answer:
We have
Percentage of people with grey hairs
The probability that the selected haired person is male :
Answer:
at most
the probability that more than
the probability that at most
.
Answer:
Total balls in urn = 25
Balls bearing mark 'X' =10
Balls bearing mark 'Y' =15
Let Z be a random variable that represents a number of balls with Y mark on them in the trial.
Z has a binomial distribution with n=6.
Answer:
Total balls in urn = 25
Balls bearing mark 'X' =10
Balls bearing mark 'Y' =15
Let Z be random variable that represents number of balls with Y mark on them in trial.
Z has binomail distribution with n=6.
Answer:
Let p and q respectively be probability that the player will clear and knock down the hurdle.
Let X represent random variable that represent number of times the player will knock down the hurdle.
Answer:
Probability of 6 in a throw of die =P
Probability that 2 sixes come in first five throw of die :
Probability that third six comes in sixth throw :
Question:8 If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
Answer:
In a leap year, there are 366 days.
In 52 weeks, there are 52 Tuesdays.
The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.
The remaining 2 days can be :
1. Monday and Tuesday
2. Tuesday and Wednesday
3. Wednesday and Thursday
4. Thursday and Friday
5.friday and Saturday
6.saturday and Sunday
7.sunday and Monday
Total cases = 7.
Favorable cases = 2
Probability of having 53 Tuesday in a leap year = P.
Answer:
Probability of success is twice the probability of failure.
Let probability of failure be X
then Probability of success = 2X
Sum of probabilities is 1.
Let
Let X be random variable that represent the number of success in six trials.
Answer:
Let the man toss coin n times.
Probability of getting head in first toss = P
The minimum value to satisfy the equation is 4.
The man should toss a coin 4 or more times.
Answer:
In a throw of die,
probability of getting six = P
probability of not getting six = q
There are three cases :
1. Gets six in the first throw, required probability is
The amount he will receive is Re. 1
2.. Does not gets six in the first throw and gets six in the second throw, then the probability
The amount he will receive is - Re.1+ Re.1=0
3. Does not gets six in first 2 throws and gets six in the third throw, then the probability
Amount he will receive is -Re.1 - Re.1+ Re.1= -1
Expected value he can win :
Question:12(i) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
Answer:
'
Let R be the event of drawing red marble.
Let
Total marbles = 40
Red marbles =15
Probability of drawing red marble from box A is
Question:12(ii) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
Answer:
Let R be event of drawing red marble.
Let
Total marbles = 40
Red marbles =15
Probability of drawing red marble from box B is
Question:12(iii) Suppose we have four boxes A,B,C and D containing coloured marbles as given below:
Answer:
Let R be event of drawing red marble.
Let
Total marbles = 40
Red marbles =15
Probability of drawing red marble from box C is
Answer:
Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted
the drug prescription.
the probability that the patient followed a course of meditation and yoga is
Answer:
Total number of determinant of second order with each element being 0 or 1 is
The values of determinant is positive in the following cases
Probability is
Answer:
Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.
Let A be the event that ball drawn is red.
When a red ball is transfered from bag 1 to bag 2.
When a black ball is transfered from bag 1 to bag 2.
Question:17 If A and B are two events such that
Choose the correct answer of the following:
Answer:
A and B are two events such that
Option A is correct.
The conditional probability of an event E, given the occurrence of the event F is given by |
If you are looking for probability class 12 ncert solutions of exercise then they are listed below.
Generally, two questions( 8 marks) are asked from this chapter in 12th board final examination. You can score these 8 marks very easily with the help of probability Class 12 ncert solutions chapter 13.
In the NCERT textbook there are 37 solved examples are given, so you can understand the concept easily. In this chapter 13 12 th class, there is a total of 81 questions in 5 exercises. You should try to solve every question given in chapter 13 class 12 maths on your own.
If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 13 probability. These probability class 12 NCERT questions are solved and explained in a step-by-step manner, so it can be understood very easily.
This chapter 13 12 th class requires lots of practice to understand it better. So, you are advised to solve miscellaneous exercise of probability class 12 ncert solutions.
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
Comprehensive explanations: The probability ncert class 12 solutions provided in NCERT Solutions for Class 12 Maths Chapter 13 Probability are explained in a comprehensive and step-by-step manner. This helps students to understand the concepts better and makes it easy for them to solve similar problems.
Easy to understand: The probability solutions class 12 are written in simple language, making it easy for students to understand and learn the concepts. The solutions are designed to cater to the needs of students of all learning levels.
Covers all the topics: The class 12 probability solutions cover all the topics in Chapter 13 Probability of Class 12 Maths. This helps students to have a comprehensive understanding of the chapter.
Exercise-wise solutions: The class 12 chapter 13 maths solutions are provided exercise-wise, which helps students to focus on specific problems and concepts that they find difficult.
Examples and illustrations: The ch 13 maths class 12 solutions include examples and illustrations to explain the concepts and solutions better. These examples help students to understand the applications of probability in real-life situations.
Benefits of NCERT solutions
NCERT solutions for Class 12 Maths Chapter 13 are very easy to understand as these are prepared and explained in a detailed manner.
At the end of every chapter, there is an additional exercise called Miscellaneous exercise which is very important for you if you wish to develop a grip on the concepts. In NCERT solutions for class 12 maths chapter 13 probability, you will get solutions for miscellaneous exercise too.
These NCERT solutions for Class 12 Maths Chapter 13 PDF download are prepared with different approaches so it will give you new ways of solving the problems.
NCERT solutions for class 12 maths chapter 13 probability are prepared and explained by the experts who know how best to answer the questions in the board exam. So, it will help you to score good marks in the exam.
Happy Reading !!!
Basic probability, conditional probability, properties of conditional probability, multiplication theorem on probability, independent events, Bayes' theorem, random variables, and its probability distributions, Bernoulli trials, and Binomial distribution are important topics of this chapter.
No, CBSE doesn’t provide NCERT solutions for any class or subject. but there are so many coaching institutions which provide solutions freely. if you are interested then you can download these from careers360 official website.
Students consider integration and application of integration as the most difficult chapters in CBSE class 12 maths but with rigorous practice, you will get conceptual clarity and will be able to have a strong grip on them also.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
Re-evaluate Your Study Strategies:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
We endeavor to keep you informed and help you choose the right Career path. Sign in and access our resources on Exams, Study Material, Counseling, Colleges etc.