# NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

**NCERT exemplar Class 12 Maths solutions chapter 13** deals with the possibility of occurrence of any event or explains how likely it is for an event to occur. The possibility of occurring or not occurring of any event cannot be predicted but can be displayed in the form of probability with the absolute level of certainty. Probability of any event is a number between 0 to 1 where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of happening of an event certainly. NCERT exemplar Class 12 Maths solutions chapter 13 help students to interrelate such knowledge with real-life problems and display the application of such knowledge in different lives scenarios and situations along with enhancing decision-making at an efficient level.

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**Also, read -** NCERT Class 12 Maths Solutions

Question:1

Answer:

Given

For a loaded die –

P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3

The die is thrown twice and

- Event of the same number each time
- Event of total score of 10 or more.

Therefore, for A,

For B,

B= EVENT OF TOTAL SCORE IS 10 OR MORE

B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

For the probability of intersection B i.e. both the events occur simultaneously,

Therefore,

P (A B) = P (5,5) + P (6,6)

P (A B) = P (5) × P (5) + P (6) ×P (6)

P (A B) = 0.1×0.1+ 0.1×0.1

P (A B) = 0.01+0.01

P (A B) = 0.02

Knowing that if two events are independent, then,

Therefore,

Hence, A and B are independent events.

Question:2

Answer:

Given

Therefore,

And

Therefore,

Therefore,

Therefore,

P (A B) ≠ P(A). P(B)

Hence, A and B are not independent events.

Question:3

Answer:

Given-

At least one of the two events A and B occurs is 0.6 i.e. P(AB) = 0.6

If A and B occur simultaneously, the probability is 0.3 i.e. P(AB) = 0.3

It is known to us that

P(AB) = P(A)+ P(B) – P(AB)

0.6 = P(A)+ P(B) – 0.3

P(A)+ P(B) = 0.6+ 0.3 = 0.9

To find-

Therefore,

Question:4

Answer:

Given

A bag contains 5 red marbles and 3 black marbles

If the first marble is red, the following conditions have to be followed for at least one marble to be black.

- Second marble is black and third is red =
- Second and third, both marbles are black =

(iii) Second marble is red and third marble is black =

Let event = drawing red marble in draw

And event = drawing black marble in draw

Hence, the for the probability for at least one marble to be black is P

Question:5

Answer:

Given-

Two dice are drawn together i.e. n(S)= 36

S is the sample space

E = a of total 4

F= a total of 9 or more

G= a total divisible by 5

Therefore, for E,

E = a of total 4

∴E = {(2,2), (3,1), (1,3)}

∴n(E) = 3

For F,

F= a total of 9 or more

∴ F = {(3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (6,5), (6,6), (5,5), (5,6)}

∴n(F)=10

For G,

G = a total divisible by 5

∴ G = {(1,4), (4,1), (2,3), (3,2), (4,6), (6,4), (5,5)}

∴ n(G) = 7

Here, (E F) = φ AND (E G) = φ

Also, (F G) = {(4,6), (6,4), (5,5)}

n (F G) = 3 and (E F G) = φ

Therefore,

P (F G) ≠ P(F). P(G)

Hence, there is no pair which is independent

Question:6

Explain why the experiment of tossing a coin three times is said to have binomial distribution.

Answer:

Let p= events of failure and q=events of success

It is known to us that,

A random variable X (=0,1, 2,…., n) is said to have Binomial parameters n and p, if its probability distribution is given by

Therefore, in the experiment of a coin being tossed three times,

we have random variable X which can take values 0,1,2 and 3 with parameters n=3 and

Hence, tossing of a coin 3 times is a Binomial distribution.

Question:7

A and B are two events such that

Find:

(i) P(A|B) (ii) P(B|A) (iii) P(A’|B) (iv) P(A’|B’)

Answer:

Given-

iv)

Question:9

By De Morgan's laws

Knowing that

Hence, either

(iv)

Hence, either

Question:11

Answer:

It has to be proven that

As we know,

Therefore,

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.

Since () means A and B both occurring at the same time and

() means A and both occurring at the same time.

Therefore, it is not possible that occur at the same time.

Hence, are mutually exclusive.

∴ P[(A B) (A B’)] = 0 ….. (1)

Therefore, A = A (B B’)

Knowing that P (A B) = P (A) + P(B) - P (A B)

Hence proved.

ii) It is to be proven that,

AB means the all the possible outcomes of both A and B.

From the above Venn diagram,

When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.

Since (A B) means A and B both occurring at the same time and

(A B’) means A and B’ both occurring at the same time.

(A' B) means A’ and B both occurring at the same time.

Therefore, it is not possible that (A B), (A B’) and ( B) occur at the same time.

Hence these events are mutually exclusive.

It is known that,

Therefore,

From (1),(2),(3) and (4) we get,

Hence Proved

Question:12

If X is the number of tails in three tosses of a coin, determine the standard deviation of X.

Answer:

Given-

Radom variable X is the member of tails in three tosses of a coin

Therefore, X= 0,1,2,3

Substituting the values of in equation (i), we get:

And standard deviation of

Question:13

Answer:

Let X = the random variable of profit per throw

Probability of getting any number on dice is .

Since, she loses Rs 1 on getting any of 2, 4 or 5.

Therefore, at X= -1,

P(X) = P (2) +P(4) +P(5)

In the same way, =1 if dice shows of either 1 or 6 .

and at X=4 if die shows a 3

Question:14

Answer:

Since three dice are thrown at the same time, the sample space is [n(S)] = 6^{3}= 216.

Let E_{1} be the event when the sum of numbers on the dice was six and

E_{2} be the event when three twos occur.

Question:15

Answer:

Let X be the variable for the prize

The possibility is of winning nothing, Rs 500, Rs 2000 and Rs 3000.

So, X will take these values.

Since there are 3 third prizes of 500, the probability of winning third prize is .

1 first prize of 3000, so probability of winning third prize is .

1 second prize of 2000, so probability of winning third prize is .

Question:16

Answer:

Given-

= [4 white balls] and = [5 black balls]

= [9 white balls] and = [7 black balls]

Let be the event that the ball transferred from the first bag is white and

be the event that the ball transferred from the bag is black.

E is the event that the ball drawn from the second bag is white.

Question:17

Answer:

Given-

Bag I= [3Black, 2White], Bag II= [2 black, 4 white]

Let E_{1} be the event that bag I is selected

E_{2} be the event that bag II is selected

E_{3} be the event that a black ball is selected

Therefore,

Question:18

Answer:

Given-

The box has 5 blue and 4 red balls.

Let E_{1} be the event that first ball drawn is blue

E_{2 }be the event that the first ball drawn is red and

E be the event that second ball drawn is blue.

Question:19

Answer:

Let E_{1}, E_{2}, E_{3 }and E_{4} be the events that the first, second, third and fourth card is king respectively.

As we know, there are 4 kings,

when 1 king is taken out, there are 3 kings and total 51 cards left.

Therefore, probability of drawing a king when one king has been taken out is:

When 2 kings are taken out, there are 2 kings and 50 cards left. Therefore, probability of drawing a king when two kings have been taken out is:

When 3 kings are taken out, 1 king and 49 cards are left.

Therefore, probability of drawing a king when three kings have been taken out is:

Probability that all 4 cards are king is:

Question:20

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

Answer:

Given-

n=5, Odd numbers = 1,3,5

and

also, r=3

Question:21

Ten coins are tossed. What is the probability of getting at least 8 heads?

Answer:

Let X = the random variable for getting a head.

Here, n=10, r≥8

r=8,9,10

Question:22

Answer:

Given-

The man shoots 7 times, therefore, n=7

And probability of hitting target is

Question:23

Answer:

Given-

There are 10 defective watches in 100 watches

The probability of defective watch from a lot of 100 watch

As we know that

Question:24

Consider the probability distribution of a random variable X:

Calculate (i) (ii) Variance of X.

Answer:

Given-

Question:25

The probability distribution of a random variable X is given below:

(i) Determine the value of k.

(ii) Determine P (X ≤ 2) and P (X > 2)

(iii) Find P (X ≤ 2) + P (X > 2).

Answer:

Given-

(i) since,

(ii)

Question:26

For the following probability distribution determine standard deviation of the random variable X.

Answer:

Given-

Question:27

Answer:

Given-

X= number of four seen

On tossing to die, X=0,1,2

Therefore,

Thus, the table is derived

Question:28

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

Answer:

Given-

X= no. of twos seen

Therefore, on throwing a die three times, we will have X=0,1,2,3

Question:29

Answer:

Given-

For the second die

Let x be the number of ones seen

For X=0,

Question:30

Two probability distributions of the discrete random variable X and Y are given below.

Prove that .

Answer:

To prove that-

Taking LHS of equation (i), we have:

Taking RHS of equation (i) we get:

Thus, from equations (ii) and (iii), we get:

Hence proved.

Question:31

Answer:

Let X be the random variable which denotes that the bulb is defective.

And

(j) None of the bulbs is defective i.e., r=0

(ii)Exactly two bulbs are defective i.e., r=2

Question:32

Answer:

Let E_{1} be the event that a fair coin is drawn

E_{2} be the event that two headed coin is drawn

E be the event that tossed coin get a head

Question:33

Answer:

Given-

Let E_{1} be the event that the person selected is of group O

E_{2} be the event that the person selected is of other than blood group O

And E_{3} be the event that the person selected is left handed

∴P(E_{1}) =0.30, P(E_{2}) =0.70

P(E_{3}|E_{1}) = 0.060 And P(E_{3}|E_{2}) =0.10

Using bayes’ theorem, we have:

Question:34

Answer:

The notation P [r ≤ p| s ≤ p] means that

P (r ≤p) *given that s ≤ p

Since we know s ≤ p , then it means that s is drawn first.

Let us have n numbers before s is drawn:

(1 . . s …. . p . . .. n)

After s is drawn,

[ 1 ... p] has one element missing, so there are (p-1) elements.

Also, there is one element missing from the entire set, so there are (n-1) altogether.

Among (1 . . s …. . p) the probability of drawing s is .

is the probability that

Therefore,

Question:35

Answer:

Let X be the random variable score obtained when a die is thrown twice.

∴ X= 1,2,3,4,5,6

The sample space is

In the same way,

Therefore, the required distribution is,

Question:36

Answer:

Given-

X=0,1,2 and P(X) at X=0 and 1,

Let X=2, P(X) is x.

The following distribution is obtained

And

And, as we know that

Question:38

Answer:

Given-

A and B throw a pair of dice alternately.

A wins if he gets a total of 6

And B wins if she gets a total of 7

Therefore,

A = {(2,4), (1,5), (5,1), (4,2), (3,3)} and

B = {(2,5), (1,6), (6,1), (5,2), (3,4), (4,3)}

Let P(B) be the probability that A wins in a throw

And P(B) be the probability that B wins in a throw

The probability of A winning the game in the third row

Question:39

Answer:

Given-

A= {(x, y):x+y=11}

And B= {(x, y): x≠5}

∴ A = {(5,6), (6,5)}

B= {(1,1), (1,2), (1,3), (1,4), ((1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Question:40

Answer:

Given-

An urn contains m white and n black balls.

Let E_{1} be the first ball drawn of white colour

E_{2} be the first ball drawn of black colour

And E_{3} be the second ball drawn of white colour

Using the probability theorem, we get,

Hence, the probability of drawing a white ball does not depend on k.

Question:41

Answer:

Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 are selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

The probability that bag i will be chosen and a ball is selected from it is i|6.

The Law of Total Probability:

In a sample space S, let E_{1},E_{2},E_{3}…….E_{n} be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E_{1},E_{2},E_{3}…….E_{n}, then

(i) Let “E” be the event that a red ball is selected.

P(E|E1) is the probability that red ball is chosen from the bag 1.

P(E|E2) is the probability that red ball is chosen from the bag 2.

P(E|E3) is the probability that red ball is chosen from the bag 3.

Therefore,

As red ball can be selected from Bag 1, Bag 2 and Bag 3.

Therefore, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.

From the law of total probability,

Let F be the event that a white ball is selected.

Therefore, P(F|E1) is the probability that white ball is chosen from the bag 1.

P(F|E2) is the probability that white ball is chosen from the bag 2.

P(F|E3) is the probability that white ball is chosen from the bag 2.

P(F|E1) = 0

As white ball can be selected from Bag 1, Bag 2 and Bag 3

Therefore, sum of individual probabilities of choosing the red from the given bags is the probability of choosing a white ball.

Question:42

Answer:

Referring to the previous question, using Bayes theorem, we get

Let E_{1}, E_{2}, and E_{3} be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

The probability that bag i will be chosen and a ball is selected from it is i|6.

Let F be the event that a white ball is selected. Therefore, is the probability that white ball is chosen from the bag 1 .

is the probability that white ball is chosen from the bag 2 .

To find: the probability that if white ball is selected, it is selected from:

(i) Bag 2

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

(ii)Bag 3

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

Using Bayes' theorem, we get the probability of as:

Question:43

Given-

A1, A2, and A3 denote the three types of flower seeds and A1: A2: A3 = 4: 4 : 2

Therefore, Total outcomes = 10

Therefore,

Let E be the event that a seed germinates and E’ be the event thata seed does not germinate.

P(E|A1) is the probability that seed germinates when it is seed A1.

P(E’|A1) is the probability that seed will not germinate when it is seed A1.

P(E|A2) is the probability that seed germinates when it is seed A2.

P(E’|A2) is the probability that seed will not germinate when it is seed A2.

P(E|A3) is the probability that seed germinates when it is seed A3.

P(E’|A3) is the probability that seed will not germinate when it is seed A3.

The Law of Total Probability:

In a sample space mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with then

(i) Probability of a randomly chosen seed to germinate.

It can be either seed A, B or C, therefore,

From law of total probability,

(ii) that it will not germinate given that the seed is of type A3

Knowing that P(A) + P(A’) =1

P(E’|A3) = 1 – P(E|A3)

(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:44

Answer:

Let events E1, E2 be the following events-

E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA

Let E be the event that on the letter, two consecutive letters TA are visible.

Since, the letter has come either from CALCUTTA or TATA NAGAR

We get the following set of possible consecutive letters when two consecutive letters are visible in the case of TATA NAGAR

{TA, AT, TA, AN, NA, AG, GA, AR}

We get the following set of possible consecutive letters in the case of CALCUTTA,

{CA, AL, LC, CU, UT, TT, TA}

Therefore, P(E|E1) is the probability that two consecutive letters are visible when letter came from TATA NAGAR

P(E|E2) is the probability that two consecutive letters are visible when letter came from CALCUTTA

To find- the probability that the letter came from TATA NAGAR.

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:45

Answer:

Given-

There are 2 bags-

Bag 1: 3 black and 4 white balls

Bag 2: 4 black and 3 white balls

Therefore, Total balls = 7

Let events E1, E2 be the following:

E1 and E2 be the events that bag 1 and bag 2 are selected respectively

We know that a die is thrown.

Therefore, total outcomes = 6

The Law of Total Probability:

In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then

P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)

Let “E” be the event that black ball is chosen.

P(E|E1) is the probability that black ball is chosen from the bag 1.

P(E|E2) is the probability that black ball is chosen from the bag 2.

Therefore,

Therefore, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.

From the law of total probability,

Question:46

Answer:

Given-

There are 3 urns U1, U2 and U3

Let U1 be 2 white and 3 black balls

U2 be 3 white and 2 black balls

U3 be 4 white and 1 black balls

Therefore, Total balls = 5

As there is an equal probability of each urn being chosen

Let be the event that a ball is chosen from an urn

Let A be the event that white ball is drawn.

P(A|E1) is the probability that white ball is chosen from urn U1

P(A|E2) is the probability that white ball is chosen from urn U2

P(A|E3) is the probability that white ball is chosen from urn U3

To find- the probability that the ball is drawn was from

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

Question:47

Answer:

Let events E1, E2, E3 be the following events:

E1 - the event that person has TB and E2 - the event that the person does not have TB

Therefore, Total persons = 1000

Therefore,

Let E be the event that the person is diagnosed to have TB

To find- the probability that the person actually has TB

Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.

is the probability that person actually has TB

Question:48

Answer:

Let E1, E2, E3 be the following events:

E1 -event that item is manufactured by machine A

E2- event that item is manufactured by machine B

E3- event that item is manufactured by machine C

Since, E1, E2 and E3 are mutually exclusive and exhaustive events hence, they represent a partition of sample space.

As we know that

Items manufactured on machine A = 50%

Items manufactured on machine B = 30%

Items manufactured on machine C = 20%

Therefore,

Let E be the event that ‘an item is defective’.

Therefore, P(E|E1) is the probability of the item drawn is defective given that it is manufactured on machine A = 2%

P(E|E2) is the probability of the item drawn is defective given that it is manufactured on machine B = 2%

P(E|E3) is the probability of the item drawn is defective given that it is manufactured on machine C = 3%

Therefore,

To find- the probability that the item which is picked up is defective, it was manufactured on machine A

Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

is the probability that the item is drawn is defective and it was manufactured on machine A

Question:49

Let X be a discrete random variable whose probability distribution is defined as follows:

where k is a constant. Calculate

(i) the value of k (ii) E (X) (iii) Standard deviation of X.

Answer:

Given-

Therefore, we get the probability distribution of X as

(i) the value of k

As we know, Sum of the probabilities =1

To find: E(X)

The probability distribution of X is:

(iii) To find: Standard deviation of X

As we know,

As we know,

standard deviation of

Question:50

The probability distribution of a discrete random variable X is given as under:

Calculate :

(i) The value of A if E(X) = 2.94

(ii) Variance of X.

Answer:

i ) Given-

E(X) = 2.94

It is known to us that μ = E(X)

x1

Question:51

The probability distribution of a random variable x is given as under:

where k is a constant. Calculate

(i) E(X) (ii) (iii) P(X ≥ 4)

Answer:

Given-

As we know, Sum of the probabilities =1

(i) To find:

or

(ii) To find:

We first find

As we know that,

Question:52

Answer:

Given-

n coins have head on both the sides and (n + 1) coins are fair coins

Therefore, Total coins = 2n + 1

Let E1, E2 be the following events:

E1 = Event that an unfair coin is selected

E2 = Event that a fair coin is selected

The Law of Total Probability:

In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then

P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)

Let “E” be the event that the toss result is a head

P(E|E1) is the probability of getting a head when unfair coin is tossed

P(E|E2) is the probability of getting a head when fair coin is tossed

Therefore,

Question:53

Answer:

Let X be a random variable of number of aces

X can take values 0, 1 or 2 because only two cards are drawn.

Therefore, Total deck of cards = 52

and total no. of ACE cards in a deck of cards = 4

Since the draws are done without replacement, therefore, the two draws are not independent.

Therefore,

P(X = 0) = Probability of no ace being drawn

= P(non – ace and non – ace)

= P(non – ace) × P(non – ace)

Question:54

Answer:

Let X be the random variable for a ‘success’ for getting an even number on a toss.

∴ X = 0, 1, 2

n = 2

Even number on dice = 2, 4, 6

∴ Total possibility of getting an even number = 3

Total number on dice = 6

p = probability of getting an even number on a toss

Question:55

Answer:

The sample space is

S = { (1,2),(1,3),(1,4),(1,5)

(2,1),(2,3),(2,4),(2,5)

(3,1),(3,2),(3,4),(3,5)

(4,1),(4,2),(4,3),(4,5)

(5,1),(5,2),(5,3),(5,4)}

Total Sample Space, n(S) = 20

Let random variable be X which denotes the sum of the numbers on the cards drawn.

∴ X = 3, 4, 5, 6, 7, 8, 9

At X = 3

The cards whose sum is 3 are (1,2), (2,1)

At x=4

The cards whose sum is 4 are (1,3),(3,1)

At X=5

The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)

At X=6

The cards whose sum is 6 are (1,5),(2,4),(4,2),(5,1)

At x=7

The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)

At X = 8

The cards whose sum is 8 are (3,5), (5,3)

At X = 9

The cards whose sum is 9 are (4,5), (5,4)

Question:56

If , and , then P(B | A) is equal to

A. 1|10

B. 1|8

C. 7|8

D. 17|20

Answer:

Given- , and

Hence, Correct option is C

Question:57

If P(A ∩ B) = 7|10 and P(B) = 17|20, then P(A|B) equals

A. 14|17

B. 17|20

C. 7|8

D. 1|8

Answer:

Given-

Hence, Correct option is A

Question:58

If , then P(B|A) + P(A|B) equals

A. 1|4

B. 1|3

C. 5|12

C. 7|2

Answer:

Given-

Hence, the Correct option is D

Question:59

If , the P(A′|B′).P(B′|A′) is equal to

A. 5|6

B. 5|7

C. 25|42

D. 1

Answer:

Given-

Hence, the correct option is C

Question:60

If A and B are two events such that , the P(A′ ∩ B′) equals

A. 1|12

B. 3|4

C. 1|4

D. 3|16

Answer:

Given-

As we know,

P(A|B) × P(B) = P(A ∩ B) [Property of Conditional Probability]

Hence, the correct option is C

Question:61

If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then P(A ∪ B) is equal to

A. 0.24

B. 0.3

C. 0.48

D. 0.96

Answer:

Given-

P(A) = 0.4, P(B) = 0.8 and

It is known that,

Hence,

Question:62

If A and B are two events and A ≠ θ, B ≠ θ, then

A. P(A | B) = P(A).P(B)

B.

C. P(A | B).P(B | A)=1

D. P(A | B) = P(A) | P(B)

Answer:

Given-

CASE 1 : If we take option

CASE 2 : If we take option (B) i.e.

this is true, knowing that this is conditional probability.

CASE 3: If we take option

Hence, the correct option is B

Question:63

A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals

A. 2|3

B. 1|2

C. 3|10

D. 1|5

Answer:

Given-

Hence, the correct option is D

Question:64

You are given that A and B are two events such that P(B)= 3|5, P(A | B) = 1|2 and P(A ∪ B) = 4|5, then P(A) equals

A. 3|10

B. 1|5

C. 1/2

D. 3|5

Answer:

Given

Hence, the correct option is C

Question:65

In Exercise 64 above, P(B | A′) is equal to

A. 1|5

B. 3|10

C. 1|2

D. 3|5

Answer:

Referring to the above solution,

Hence, the correct option is D

Question:66

If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) =

A. 1|5

B. 4|5

C. 1|2

D. 1

Answer:

Given-

Hence, the correct option is D

Question:67

Let P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13. Then P(A′|B) is equal to

A. 6/13

B. 4/13

C. 4/9

D. 5/9

Answer:

Given-

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

Question:68

If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’|B’) equals

A. 1 – P(A|B)

B. 1 – P (A’|B)

C.

D. P(A’) | P(B’)

Answer:

Given-

By de Morgan's Law:

Option c is correct answer.

Question:69

If A and B are two independent events with P(A) = 3/5 and P(B) = 4/9, then P (A′ ∩ B′) equals

A.4/15

B. 8/45

C. 1/3

D. 2/9

Answer:

Question:70

If two events are independent, then

A. they must be mutually exclusive

B. the sum of their probabilities must be equal to 1

C. (A) and (B) both are correct

D. None of the above is correct

Answer:

Events which cannot happen at the same time are known as mutually exclusive events. For example: when tossing a coin, the result can either be heads or tails but cannot be both.

Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).

Eg: Rolling a die and flipping a coin. The probability of getting any number on the die will not affect the probability of getting head or tail in the coin.

Therefore, if A and B events are independent, any information about A cannot tell anything about B while if they are mutually exclusive then we know if A occurs B does not occur.

Therefore, independent events cannot be mutually exclusive.

To test if probability of independent events is 1 or not:

Let A be the event of obtaining a head.

P(A) = 1/2

Let B be the event of obtaining 5 on a die.

P(B) = 1/6

Now A and B are independent events.

Hence option D is correct.

Question:71

Let A and B be two events such that P(A) = 3/8, P(B) = 5/8 and P(A ∪ B) = 3/4. Then P(A | B).P(A′ | B) is equal to

A.2/5

B. 3/8

C. 3/20

D. 6/25

Answer:

Option D is correct.

Question:72

If the events A and B are independent, then P(A ∩ B) is equal to

A. P (A) + P

B. (B) P(A) – P(B)

C. P (A) . P(B)

D. P(A) | P(B)

Answer:

We know that if events A and B are independent,

From the definition of the independent Event,

Option C is correct.

Question:73

Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P(E | F)–P(F | E) equals

A. 2/7

B. 3/25

C. 1/70

D. 1/7

Answer:

Given-

P(E) = 0.3, P(E ∪ F) = 0.5

Also, E and F are independent, therefore,

P (E ∩ F)=P(E).P(F)

As we know , P(E ∪ F)=P(E)+P(F)- P(E ∩ F)

P(E ∪ F)=P(E)+P(F)- [P(E) P(F)]

Question:74

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

A. 45/196

B. 135/392

C. 15/56

D. 15/29

Answer:

The Probability of getting exactly one red ball is

P(R).P(B).P(B) + P(B).P(R).P(B) + P(B).P(B).P(R)

Option C is correct.

Question:75

Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is

A. 1/3

B. 4/7

C. 15/28

D. 5/28

Answer:

Given-

A bag contains 5 red and 3 blue balls

Therefore, Total Balls in a Bag = 8

For exactly 1 red ball probability should be

3 Balls are drawn randomly then possibility for getting 1 red ball

P(E)=P(R).P(B)+P(B).P(R)

Option B is correct.

Question:76

Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is

A. 0.024

B. 0.188

C. 0.336

D. 0.452

Answer:

Given-

Hence, Probability of two hits is 0.188

Option B is correct.

Question:77

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

A. 1/2

B. 1/3

C. 2/3

D. 4/7

Answer:

The statement can be arranged in a set as S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}

Let A be Event that a family has at least one girl, therefore,

A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)

Let B be Event that eldest child is girl then, therefore,

B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)

(A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)

since,

Hence,

Option D is correct.

Question:78

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

A. 1/2

B. 1/4

C. 1/8

D. 3/4

Answer:

Let A be Event for getting number on dice and B be Event that a spade card is selected

Therefore,

A={2,4,6}

B={13}

Question:79

Answer:

Given-

There are total 8 balls in box.

Therefore, P(G) , Probability of green ball

P(B) , Probability of blue ball

The probability of drawing 2 green balls and one blue ball is

P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)

Question:80

Answer:

Given-

Total number of batteries: n= 8

Number of dead batteries are = 3

Therefore, Probability of dead batteries is

If two batteries are selected without replacement and tested

Then, Probability of second battery without replacement is

Required probability =

Question:81

Eight coins are tossed together. The probability of getting exactly 3 heads is

A.

B.

C.

D.

Answer:

Given-

probability distribution

Total number coin is tossed, n=8

The probability of getting head,

The probability of getting tail,

The Required probability

Question:82

Answer:

Let A be the event that the sum of numbers on the dice was less than 6

And B be the event that the sum of numbers on the dice is 3

Therefore,

A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)

n(A)=10

B={(1,2)(2,1)

n(B)=2

Required probability =

Required probability =

Hence, the probability is

Question:83

Which one is not a requirement of a binomial distribution?

A. There are 2 outcomes for each trial

B. There is a fixed number of trials

C. The outcomes must be dependent on each other

D. The probability of success must be the same for all the trials

Answer:

In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.

Hence option c is correct.

Question:84

Answer:

We know that

Number of cards = 52

Number of queens = 4

Therefore, Probability of queen out of 52 cards =

According to the question,

If a deck of card shuffled again with replacement, then

Probability of getting queen is ,

Therefore, The probability, that both cards are queen is ,

Hence, Probability is

Question:85

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

A.

B.

C.

D.

Answer:

We know that in the examination, we have only two option True and False.

Therefore, the probability of getting True is, p= 1/2

And, probability of getting False is, q=1/2

The total number of Answer in examination, n=10

The probability of guessing correctly at least 8 it means r=8,9,10

As we know, the probability distribution

Hence, Probability is

Question:86

Answer:

Given-

Total number of person n=5

Total number of swimmers among total person , r=4

Probability of not swimmer , Q=0.3

Therefore, The probability of swimmer , p=1-Q=0.7

As we know that the probability distribution

=

Hence,

Option A is correct.

Question:87

The probability distribution of a discrete random variable X is given below:

The value of k is

A. 8

B. 16

C. 32

D. 48

Answer:

Given-

Probability distribution table

As we know

Hence, the value of k is 32

Option C is correct.

Question:88

For the following probability distribution:

E(X) is equal to:

A. 0

B. –1

C. –2

D. –1.8

Answer:

Given-

Probability distribution table

Option D is correct.

Question:89

For the following probability distribution

is equal to

A. 3

B. 5

C. 7

D. 10

Answer:

Given-

Probability distribution table

Option D is correct.

Question:90

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If P(x = r) / P(x = n–r) is independent of n and r, then p equals

A. 1/2

B. 1/3

C. 1/5

D. 1/7

Answer:

According to the question, this expression is independent of n and r if

Hence

Option A is correct.

Question:91

Answer:

Let A be the event that students failed in physics.

As per the question, 30% students failed in physics.

∴ P(A) = 0.30

Similarly, if we denote the event of failing in maths with B.

We get P(B) = 0.25

And probability of failing in both subjects can be represented using intersection as

P (A ∩ B) = 0.1

To find- a conditional probability of failing of student in physics given that she has failed in mathematics.

The situation can be represented mathematically as-

P(A|B) =?

Using the fundamental idea of conditional probability, we know that:

Our answers clearly match with option B

is the only correct choice.

Question:92

Answer:

Let E be the event that student ‘A’ solves the problem correctly.

∴ P(E) = 1/3

In the same way if we denote the event of ’B’ solving the problem correctly with F

We get P(F) = 1/4

Since both the events are independent.

∴ Probability that both the students solve the question correctly can be represented as-

Probability that both the students could not solve the question correctly can be represented as-

Given: probability of making a common error and both getting same answer.

If they are making an error, we can be sure that answer coming out is wrong.

Let S be the event of getting same answer.

above situation can be represented using conditional probability.

And if their answer is correct obviously, they will get same answer.

To find- the probability of getting a correct answer if they committed a common error and got the same answer.

Mathematically, i.e,

By observing our requirement and availability of equations, we can use Bayes theorem to solve this.

Using Bayes theorem, we get-

Substituting the values from above -

Our answers clearly match with option D.

∴ Option (D) is the only correct choice.

Question:93

Answer:

We can solve this using Bernoulli trials.

Here n = 5 (as we are drawing 5 pens only)

Success is defined when we get a defective pen.

Let p be the probability of success and q probability of failure.

∴ p = 10/100 = 0.1

And q = 1 – 0.1 = 0.9

To find- the probability of getting at most 1 defective pen.

Let X be a random variable denoting the probability of getting r number of defective pens.

∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)

The binomial distribution formula is:

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

Our answer matches with option D.

∴ Option (D) is the only correct choice.

Question:94

Answer:

FALSE

Events are mutually exclusive when–

P(A∪B) = P(A) + P(B)

But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that

P(A ∩ B) ≠ 0

Events are independent when–

P(A ∩ B) = P(A)P(B)

Again P(A) > 0 and P(B)> 0 are not sufficient conditions to validate them.

Question:95

Answer:

TRUE

As A and B are independent

hence proved

Question:96

Answer:

False

If A and B are mutually exclusive, that means

P(A∪B) = P(A) + P(B)

From this equation it cannot be proved that

P(A ∩ B)= P(A)P(B).

Hence, it is a false statement.

Question:97

Answer:

False

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)

From the equation it cannot be proved that

P(A∪B) = P(A) + P(B)

It is only possible if either P(A) or P(B) = 0, which is not given in question.

Hence, it is a false statement.

Question:98

Answer:

TRUE

If A and B are independent events it means that

P(A ∩ B) = P(A)P(B)

Thus, from the definition of independent event we say that statement is true.

Question:99

State True or False for the statements in the Exercise.

Another name for the mean of a probability distribution is expected value.

Answer:

TRUE

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

Question:100

Answer:

TRUE

If A and B are independent events, it means that

P(A ∩ B) = P(A)P(B)

P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)

and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.

Hence Proved

Question:101

Answer:

TRUE

If A and B are independent events, that means

∴ Statement is true.

Question:102

Answer:

True

The above equation means that:

we need to add to get the equal term

LHS is greater.

Question:103

Answer:

True

Let A, B,C be the occurrence of events A,B and C and A’,B’ and C’ not occurrence.

P(A) = P(B) = P(C) = p and P(A’) = P(B’) = P(C’) = 1-p

P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C) + P(A ∩ B ∩ C)

events are independent:

Hence, statement is true.

Question:106

Answer:

p = 1/10

As n = 5 {representing no. of trials}

p = probability of success

As it is a binomial distribution.

∴ probability of failure = q = 1 – p

Given-

P(X = 2) = 9.P(X = 3)

The binomial distribution formula is:

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

using binomial distribution,

Question:107

Answer:

Variance is the mean of deviation of Random variable from its expected value.

On expanding we get the formula:

Question:108

Answer:

we know that

Given-

This implies that A and B are independent of each other.

NCERT exemplar solutions for Class 12 Maths chapter 13 provided here for the NCERT books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.

NCERT exemplar Class 12 Maths solutions chapter 13 PDF download are also available for students for extended learning. The topics covered are as follows:

**Main Subtopics**

- Introduction
- Conditional Probability
- Properties of conditional probability
- Multiplication theorem on probability
- Independent events
- Bayes’ Theorem
- Partition of a sample space
- Theorem of total probability
- Random Variables and its Probability Distributions
- Probability distribution of a random variable
- Mean of a random variable
- Variance of a random variable
- Bernoulli Trials and Binomial Distribution
- Bernoulli trials
- Binomial distribution

**What will the students learn from NCERT Exemplar Class 12 Math Solutions Chapter 13?**

Today’s society has adopted a very practical approach to life and has understood the principle of chance in nature. Similarly, the importance of studying probability could be inferred from the basic lives of people which displays a great example at how there is a constant need to understand the possibility and extract information out of it for one’s benefit. NCERT exemplar Class 12 Math solutions chapter 13 will help students to understand each topic in a better way. Such instances of life display the necessity and importance of learning the concept of probability or possibility of any event to make important decisions in their life and how they are drawn based on chances. Such as in life you come across a lot of data which need to be interpreted, and decisions are to be extracted out of such data on the basis of probable thinking which necessitates the learning of probabilistic thinking.

NCERT exemplar Class 12 Math solutions chapter 13 are easy to grasp and can be helpful in scoring well in the exams. It is not only theoretical/academic oriented but provides a great exposure regarding the nature of chance and variation in life when you encounter such situations in real life. It also develops and exhibits real-life situations teaching the art of risk analysis and management, providing a better approach towards understanding risks in real life and displaying relative chances of events in one’s life. This topic also helps in scientific reasoning and also has its imprint in various fields such as engineering, mathematician, research analyst, etc. Probability also supports statistics and helps improve and analyze data-driven decision-making skills for real life.

**NCERT Exemplar Class 12 Maths Solutions**

**Important Topics To Cover From NCERT Exemplar Class 12 Math Solutions Chapter 13**

- Class 12 Math NCERT exemplar solutions chapter 13 displays important concept including that of the conditional probability of any event given that another event has already occurred which will be helpful in understanding different theorems relating to probability, multiplication rule of probability, independence of events and discrete probability distribution also known as Binomial distribution.
- NCERT exemplar Class 12 Math solutions chapter 13 also highlights the concept of the random variable, its probability distribution and also the mean and variance of any probability distribution.

**NCERT Exemplar Class 12 Solutions**

### Also, check NCERT Solutions for questions given in the book:

Chapter 1 | |

Chapter 2 | |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | |

Chapter 6 | |

Chapter 7 | |

Chapter 8 | |

Chapter 9 | |

Chapter 10 | |

Chapter 11 | |

Chapter 12 | |

Chapter 13 | Probability |

#### Must Read NCERT Solution subject wise

- NCERT Solution for Class 12 Physics
- NCERT Solution for Class 12 Chemistry
- NCERT Solution for Class 12 Maths
- NCERT Solution for Class 12 Biology

#### Read more NCERT Notes subject wise

- NCERT Notes for Class 12 Physics
- NCERT Notes for Class 12 Chemistry
- NCERT Notes for Class 12 Maths
- NCERT Notes for Class 12 Biology

#### Also Check NCERT Books and NCERT Syllabus here

## Frequently Asked Question (FAQs) - NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

**Question: **Are these questions useful in competitive exams?

**Answer: **

Yes, these NCERT exemplar Class 12 Maths solutions chapter 13 pdf download are useful in entrance exams for engineering and also are helpful for many other competitive exams.

**Question: **Are these questions helpful in board exams?

**Answer: **

Yes, these solutions are highly useful for board exams as it helps in learning the steps and the way the CBSE wants students to solve questions.

**Question: **How many questions are included in solutions?

**Answer: **

All these questions that are mentioned in the main exercise after the chapter and in the additional question section are solved in NCERT exemplar Class 12 Maths solutions chapter 13.

**Question: **How can probability be useful?

**Answer: **

Understanding probability can be highly effective in solving real life problems and also in understanding higher education topics.

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