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In this page we have *NCERT Solutions for Class 12 Maths Chapter 13: Probability* for
EXERCISE 13.1 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

Given below are events such as P(E) = 0.6, P(F) = 0.3 P(E∩F) = 0.2. Determine the value of P(E/F) and P(F|E)

Given that P(E) = 0.6, P(F) = 0.3, P (E∩F) = 0.2

P(E|F) = P(E∩F)/P(F) = 0.2/0.3 = 2/3

P(F|E) = P(F∩E)/ P(E) = 0.2/0.6 = 1/3

Determine the value of P (A|B) if A & B are the events such that P(B) = 0.5 and P(A∩B) = 0.32

Given that P(B) = 0.5 , P(A∩B) = 0.32.

P(A|B) = P(A∩B)/P(B)

= 0.32/0.5

P(A) = 0.8 , P(B) = 0.5 , P(B|A) = 0.4. Determine the value of

(a) P(A∩B)

(b) P(A|B)

(c) P(A∪B)

Given

P(A) = 0.8 , P(B) = 0.5 , P(B|A) = 0.4

= P(A)P(B|A)

Therefore, P(A∩B) = 0.32

= 0.8+0.5–32

Determine the value of P(AUB) if 2P(A) = P(B) = 5/13 & P(A|B )=2/5

2P(A) = P(B) = 5/13 & P(A|B )=25

P(A) = 5/26

P(B) = 5/13

P(A|B)=2/5

P(A∩B)/P(B) = 2/5

P(A∩B)= 2/5×5/13

= 2/13

We know that:

P(A∪B)=P(A)+P(B)–P(A∩B)

= 5/26+5/13–2/13

= 11/26

If A and B are events such that P(A) = 6/11 , P(B) = 5/11, P(A∪B) = 7/11. Determine the value of

(a) P(A∩B)

(b) P(A|B)

(c) P(B|A)

Given

P(A) = 6/11, P(B) = 5/11, P(A∪B) = 7/11.

7/11 = 6/11 + 5/11 – P(A∩B)

P(A∩B) = 11/11 –7/11

P(A∩B) = 4/11

= (4/11) / (5/11) = 4/5

= (4/11)/ (6/11) = 2/3

An experiment consists of tossing up of a coin three times. Determine the following:

(a) E: obtaining heads on third toss & F: obtaining heads from the first consecutive two tosses.

(b) E: obtaining at least two heads & F: obtaining at most two heads.

(c) E: obtaining at most two tails & F: obtaining at most one tail.

Determine P(E|F) in each of these cases

Given a coin is tossed thrice in an order to conduct an experiment.

Sample space (S) = {TTT , TTH ,THT , THH , HTT , HTH , HHT , HHH }

Number of favorable outcomes of event F = { HHT , HHH}

E∩F = { HHH }

P(E∩F) =1/8

P(F) = 2/8=1/4

P(E|F) = P(E∩F)/P(F) = (1/8) / (1/4) = 1/2

Number of favorable outcomes of event F = {TTT , TTH , THT , THH , HTT , HTH , HHT}

E∩F = { HHT , HTH , HHT }

P(E∩F) =3/8

P(F) = 7/8

P(E|F) = P(E∩F)/P(F) = (3/8) / (7/8) = 3/7

Number of favorable outcomes of event F = {TTT , TTH , THT , THH , HTT , HHT,HTH}

E∩F = {TTH , THT , THH .HTH , HTT , HHT }

P(E∩F) =6/8

P(F) = 7/8

P(E|F) = P(E∩F)/P(F)= (6/8) / (7/8) = 6/7

Two coins are tossed once, where

(a) E: obtaining tail on one coin & F: head appears in one coin

(b) E: tail does not appear & F: head does not appear

Two coins are tossed once

Sample space (S) = {TT, TH, HT, HH}

Number of favorable outcomes of event F = {TH, HT}

E∩F = { TH , HT }

P(E∩F) = 2/4= 1/2

P(F) = 2/4=1/2

P(E|F) = P(E∩F)/ P(F)

Number of favorable outcomes of event F = {TT }

E∩F = { ? }

P(E∩F) = 0

P(F) = 1/4

P(E|F) = P(E∩F)/ P(F)

A die is thrown three times,

E: the number 4 appears during the third toss

F: 6 & 5 appears consecutively during the first two tosses

Given that a die is being tossed thrice.

Total number of elements in the sample space = 6x6x6=216

Favorable outcomes of event E =

{ (6,1,4) , (6,2,4) , (6,3,4) , (6,4,4) , (6,5,4) , (6,6,4)

(5,1,4) , (5,2,4) , (5,3,4) , (5,4,4) , (5,5,4) , (5,6,4)

(4,1,4) , (4,2,4) , (4,3,4) , (4,4,4) , (4,5,4) , (4,6,4)

(3,1,4) , (3,2,4) , (3,3,4) , (3,4,4) , (3,5,4) , (3,6,4)

(2,1,4) , (2,2,4) , (2,3,4) , (2,4,4) , (2,5,4) ,(2,6,4)

(1,1,4) , (1,2,4) , (1,3,4) , (1,4,4) , (1,5,4) , (1,6,4) }

Favorable outcomes of event F = { (6,5,6) , (6,5,5) , (6,5,4) , (6,5,3) , (6,5,2) , (6,5,1) }

E∩F = { (6,5,4) }

P(F) = 6/216

P(E∩F) = 1/216

P(E|F) = P(E∩F)/P(F) = (1/216) / (6/216) = 1/6

Mother, father and son line up at random for a family picture

E: the son is placed in one of the end

F: the position of the father is in the middle

Mother, father and son line up at random for a family picture

Let us consider the father , mother and the child are denoted by F , M and S respectively.

Sample space = { SFM , SMF , FSM , FMS , MSF , MFS }

Favorable outcomes of event E = { SFM , SMF , FMS , MFS }

Favorable outcomes of event F = { SFM , MFS }

P(F) = 2/6=1/3

P(E∩F) = 2/6=1/3

P(E|F) = P(E∩F)/P(F) = (1/3) / (1/3)

A black and a red dice are rolled.

(a) Determine the conditional probability of getting a sum of numbers greater than 9 such that the black die results in a 5

(b) Determine the conditional probability of getting a sum of numbers 8 such that red die results in numbers less than 4

Given A black and a red dice are rolled.

Total number of elements in the sample space

Favorable outcomes of event F = { (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) }

E∩F = { (5,6) , (5,5) }

Conditional probability of getting a sum greater than 9 , so that the black die results in 5 .

P (E|F) = P(E∩F)/P(F)

= (2/36) / (6/36) = 1/3

b) Favorable outcomes of event E = {(2,6) , (3,5) , (4,4) , (5,3) ,(6,2) }

Favorable outcomes of event F =

{ (1,1) , (1,2) , (1,3) , (2,1) , (2,2) , (2,3) ,

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

(5,1) , (5,2) , (5,3) , (6,1) , (6,2) , (6,3)}

E∩F = { (5,3) , (6,2) }

Conditional probability of getting a sum of numbers 8 such that red die results in numbers less than 4.

P (E|F) = P(E∩F)/P(F)

= (2/36) / (18/36) = 1/9

A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}

Find

(i) P(E|F) and P (F|E)

(ii) P(E|G) and P(G|E)

(iii) P((E ∪ F)|G) and P(E ∩ F)|G)

E = {1,3,5}, F = {2,3} and G = {2,3,4,5}

Sample space = { 6 , 5 , 4 , 3 , 2 , 1}

Favorable outcomes of event E = { 5 , 3 , 1 }

Favorable outcomes of event F = { 3 , 2 }

Favorable outcomes of event G = { 5 , 4 , 3 , 2 }

Therefore, P(E) = 3/6=1/2

P(F) = 2/6=1/3

P(G) = 4/6=2/3

P (E∩F) = 1/6

P(E|F) =P(E∩F)/P(F) = (1/6) / (1/3)=1/2

P(F|E) =P(E∩F)/ P(E) = (1/6) / (1/2)=1/3

P (E∩G) = 1/6

P(E|G) =P(E∩G)/ P(G)=(1/3) / (2/3)=1/2

P(G|E) =P(E∩G)/ P(E) =(1/3) / (1/)2=2/3

(E∪F)∩G = {5,3,2}

E∩F = {3}

(E∩F)∩G= = { 3 }

P(E∪F)=4/6=2/3

P[(E∪F)∩G]=3/6=1/2

P(E∩F)=1/6

P[(E∪F)|G]=P[(E∪F)∩G]/P(G)=(1/2) / (2/3)=3/4

P[(E∩F)|G]=P[(E∩F)∩G]/P(G)=(1/6) / (2/3)=1/4

Assume that each of the children born in a family can either be a boy or a girl. If a family having 2 children, determine the conditional probability that both are girls given:

(a) The youngest child is a girl

(b) At least one is a girl.

Given that a family is having 2 children where both are girls.

Let us represent boy and the girl child with the letter (b) and (g) respectively.

Sample space = {(g, g), (g, b), (b, g), (b, b)}

Let us consider E be the event which indicates that both child born to a family are girls.

E = {(g, g)}

(a) Let us consider F be the event that the youngest child born in the family is a girl.

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

E∩F=(g,g)

P (F) = 2/4=1/2

P (E∩F)=1/4

P (E|F)=P(E∩F)/P(F)=1/2

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

E∩G=(g,g)

P(G) = 3/4

P (E∩G)=1/4

P (E|G)=P(E∩G)/P(G)=1/3

An instructor has a question bank consisting of 300 easy True/False questions, 200difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Let us consider below events

Easy Questions = E

Difficult Questions = D

Multiple choice Questions = M

False / True Questions = T

Total Questions in the Question bank = 1400

Number of multiple choice Questions = 900

Number of False / True Questions = 500

Probability of getting an easy multiple choice Question in the Question bank =

P (E∩M) =500/1400=5/14

Probability of selecting a multiple-choice Questions be it easy or difficult

P (M) = 900/1400=914

The conditional probability of selecting an easy question given that it is a multiple choice question

Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Total number of elements in the sample space = 36

Let E be the event that the sum of two different numbers is 4 & F be the event that two numbers appearing on both the faces of the dice are different.

E = {(3, 1), (2, 2), (1, 3)}

F = { (6 , 1) , (6 , 2) , (6 , 3) , (6 , 4) , (6 , 5) , (6 , 6)

(5 , 1) , (5 , 2) , (5 , 3) , (5 , 4) , (5 , 5) , (5 , 6)

(4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6)

(3 , 1) , (3 , 2) , (3 , 3) , (3 , 4) , (3 , 5) , (3 , 6)

(2 , 1) , (2 , 2) , (2 , 3) , (2 , 4) , (2 , 5) , (2 , 6)

(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (1 , 6) }

E∩F=(1,3),(3,1))

P(F) = 30/36=5/6

E∩F=2/36=1/18

Then

P(E|F) = P(E∩F)/ P(F) = (1/18) / (5/6) = 1/15

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. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Sample space of the above conducted experiment

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

Let E be the event that a tail appears

F be the event at least one die shows 3

E = {(1, T), (2, T), (4, T), (5, T)}

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

E∩F=?

So P(E∩F)=0

Then,

= (1/6) (1/6) + (1/6) (1/6) + (1/6) (1/6) + (1/6) (1/6) + (1/6) (1/6) + (1/6) (1/6) + (1/6) (1/6) = 7/36

P (E|F) = 0/(7/36) = 0

If P (A) = 12 P (B) = 0,

Then P (A|B) =?

(a) 0

(b) 12

(c) not defined

(d) 1

It is given that:

P (A) = 12

P (B) = 0

P(A|B)

= P(A∩B)/ P(B)

= not defined

Hence, the correct answer is C

If A and B are events such that P(A|B) = P(B|A) , then :

(a) A⊂B , A ≠ B

(b) A = B

(c) A∩B=?

(d) P (A) = P(B)

Given

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

P(A∩B)/P(B) = P(A∩B)/P(C)

P (A) = P(B)

The correct answer is D

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