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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.
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.
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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.
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(AB)) is defined as:
P(AB) = P(A ∩ B) / P(B) (when P(B) ≠ 0)
Other conditional probability relationships:
P(SF) = P(FF) = 1
P((A ∪ B)F) = P(AF) + P(BF) – P((A ∩ B)F)
P(E'F) = 1 − P(EF)
>> 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(FE) = P(F) P(EF) (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(EiA) = [P(Ei) * P(AEi)] / ∑[P(Ej) * P(AEj)], 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(AE1) + P(E2) * P(AE2) + … + P(En) * P(AEn)
>> Random Variables and their Probability Distributions: A random variable is a realvalued 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 probabilityExercise: 13.1
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.
NCERT solutions for class 12 maths chapter 13 probabilityExercise: 13.2
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.
NCERT solutions for class 12 maths chapter 13 probabilityExercise: 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 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.
NCERT solutions for class 12 maths chapter 13 probabilityExercise: 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
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 
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
X can be 0,1,2,3
The probability distribution is as
X  0  1  2  3 
P(X) 




mean number of heads :
Question:11 Two dice are thrown simultaneously. If denotes the number of sixes, find the expectation of .
Answer:
denotes the number of sixes, when two dice are thrown simultaneously.
X can be 0,1,2.
Not getting six on dice
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 ways.
denote the larger of the two numbers obtained.
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:
denote the sum of the numbers obtained when two fair dice are rolled.
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 nonace cards.
The probability distribution is as :
X  0  1  2 
P(X) 



Option D is correct.
NCERT solutions for class 12 maths chapter 13 probabilityExercise: 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 times.X be getting a doublet.
Probability of getting doublet in a throw of pair of dice :
X has a binomial distribution,n=4
Put x = 2
Answer:
There are defective items in a large bulk of items.
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 wellshuffled deck of cards. What is the probability that
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 wellshuffled deck of cards.
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 wellshuffled 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 wellshuffled deck of cards.
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 wellshuffled 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 wellshuffled deck of cards.
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 days of use is . Find the probability that out of such bulbs
none will fuse after days of use.
Answer:
Let X represent number of bulb that will fuse after days of use .Trials =5
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 days of use is Find the probability that out of such bulbs
not more than one will fuse after days of use.
Answer:
Let X represent a number of the bulb that will fuse after days of use. Trials =5
X has a binomial distribution,n=5.
Put ,
Question:5(iii) The probability that a bulb produced by a factory will fuse after days of use is Find the probability that out of such bulbs
more than one will fuse after days of use.
Answer:
Let X represent number of bulb that will fuse after days of use .Trials =5
X has a binomial distribution,n=5.
Put ,
Question:5(iv) The probability that a bulb produced by a factory will fuse after days of use is . Find the probability that out of such bulbs
at least one will fuse after days of use.
Answer:
Let X represent number of bulb that will fuse after days of use .Trials =5
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 Show that is the most likely outcome.
(Hint : is the maximum among all of , )
Answer:
X is a random variable whose binomial distribution is
Here , n=6 and .
is maximum if is maximum.
is maximum so for x=3 , is maximum.
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 exactly twice in throws of a die.
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 sixes in throws of a single die.
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 probabilityMiscellaneous Exercise
Question:1(i) A and B are two events such that Find if
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 of men and of women have grey hair.
Percentage of people with grey hairs
The probability that the selected haired person is male :
Answer:
of people are righthanded.
at most of a random sample of people are righthanded.
the probability that more than of a random sample of people are righthanded is given by,
the probability that at most of a random sample of people are righthanded is given by
.
Answer:
Total balls in urn = 25
Balls bearing mark 'X' =10
Balls bearing mark 'Y' =15
balls are drawn with replacement.
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
balls are drawn with replacementt.
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 and
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 respectively denote the event of selecting box A, B, C.
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 respectivly denote event of selecting box A,B,C.
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 respectivly denote event of selecting box A,B,C.
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.
and
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 and then
Choose the correct answer of the following:
Answer:
A and B are two events such that and
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 stepbystep 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 stepbystep 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.
Exercisewise solutions: The class 12 chapter 13 maths solutions are provided exercisewise, 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 reallife 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.
Here are some options you can explore to get admission in a good school even though admissions might be closed for many:
Waitlist: Many schools maintain waitlists after their initial application rounds close. If a student who secured a seat decides not to join, the school might reach out to students on the waitlist. So, even if the application deadline has passed, it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.
Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush. Reach out to the schools directly to see if they have any open seats in 10th grade.
Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.
Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.
Best CBSE schools in Delhi: Click Here
In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:
Choosing Your Stream:
Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.
Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.
Engineering Subjects (for a Degree):
Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.
HumanComputer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.
Here's why 2025 is more likely:
Looking Ahead (2025 Admissions):
Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore, with your academic background. Here's why:
High CAT Score: A score exceeding 99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes CAT scores heavily in the shortlisting process.
Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.
However, the shortlisting process is multifaceted:
Here's what you can do to strengthen your application:
Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.
Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.
Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
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