# NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

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## NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

Ex 13.1 Class 12 Maths Question 1
Given that E and Fare events such that
P (E) = 0.6, P (F) = 0.3 and P(E∩F) = 0.2
find P(E|F) and P (F|E).
Solution:
Given: P (E)=0.6, P (F)=0.3, P (E ∩ F)=0.2
$P(E|F)=frac { P(Ecap F) }{ P(E) } =frac { 0.2 }{ 0.3 } =frac { 2 }{ 3 }$
$P(F|E)=frac { P(Ecap F) }{ P(E) } =frac { 0.2 }{ 0.6 } =frac { 1 }{ 3 }$

Ex 13.1 Class 12 Maths Question 2
Compute P(A|B) if P(B)=0.5 and P (A∩B) = 0.32.
Solution:
Given: P (B)=0.5, P(A∩B)=0.32
$P(A|B)=frac { P(Acap B) }{ P(B) } =frac { 0.32 }{ 0.50 } =frac { 32 }{ 50 } =0.64$

Ex 13.1 Class 12 Maths Question 3.
If P (A)=0.8, P (B)=0.5 and P(B/A)=0.4, find
(i) P(A∩B)
(ii) P(A/B)
(iii)P(A∪B)
Solution:
(i) $P(B/A)=frac { P(Acap B) }{ P(A) } Rightarrow 0.4=frac { P(Acap B) }{ 0.8 }$
∴ P(A∩B) = 0.4 x 0.8 = 0.32
(ii) P(A/B) = $P(A/B)=frac { P(Acap B) }{ P(B) } =frac { 0.32 }{ 0.5 } =frac { 32 }{ 50 } =frac { 16 }{ 25 }$
(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98

Ex 13.1 Class 12 Maths Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = $frac { 5 }{ 13 }$ and P(A|B) = $frac { 2 }{ 5 }$.
Solution:
Given:
2P(A) = P(B) = $frac { 5 }{ 13 }$ and P(A|B) = $frac { 2 }{ 5 }$.

Ex 13.1 Class 12 Maths Question 5.
If P(A) = $frac { 6 }{ 11 }$,P(B) =$frac { 5 }{ 11 }$ and P(A∪B) = $frac { 7 }{ 11 }$,
Find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Solution:
Given:
P(A) = $frac { 6 }{ 11 }$,P(B) =$frac { 5 }{ 11 }$ and P(A∪B) = $frac { 7 }{ 11 }$,

Ex 13.1 Class 12 Maths Question 6.
Determine P(E/F) in question 6 to 9:
A coin is tossed three times, where
(i) E: head on third toss F: heads on first two tosses.
(ii) E: at least two heads F : at most two heads
(iii) E: at most two tails F: at least one tail
Solution:
(i) E = Head occurs on third toss as {HHH, HTH, THH, TTH}
F : Heads on first two tosses = {HHH, HHT} E ∩ F = {HHH}

Ex 13.1 Class 12 Maths Question 7.
Two coins are tossed once
(i) E: tail appears on one coin F: one coin shows head
(ii) E: no tail appears F: no head appears.
Solution:
S = {HH, TH, HT, TT} n (S) = 4
(i) E : tail appears on one coin
E = {TH,HT}, P{E) = $frac { 1 }{ 2 }$
F : one coin shows head,

Ex 13.1 Class 12 Maths 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.
Solution:
A die is thrown three times E : 4 appears on third toss = {(1,1,4), (1,2,4), (1,3,4), (1,4,4), (1,5,4),
(1.6.4),(2,1,4), (2,2,4), (2,3,4), (2,4,4), (2,5,4),
(2.6.4), (3,1,4), (3,2,4), (3,3,4), (3,4,4), (3,5,4),
(3.6.4), (4,1,4), (4,2,4), (4,3,4), (4,4,4), (4,5,4),
(4.6.4), (5,1,4), (5,2,4), (5,3,4), (5,4,4), (5,5,4),
(5.6.4), (6,1,4), (6,2,4), (6,3,4), (6,4,4), (6,5,4),
(6.6.4)}
These are 36 cases
F: 6 and 5 appears respectively on first two tosses = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6, 5,5), (6,5,6)}
These are six cases E ∩ F = {6,5,4}

Ex 13.1 Class 12 Maths Question 9.
Mother, father and son line up at random for a family picture:
E: son on one end, F: father in middle
Solution:
Mother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
F: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
E∩F = {(m, f, s), (s, f, m)}

Ex 13.1 Class 12 Maths Question 10.
A Mack and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
(a) n(S) = 6 x 6 = 36
Let A represent obtaining a sum greater than 9 and B represents black die resulted in a 5.
A= {46,64,55,36,63,45,54,65,56,66}

Ex 13.1 Class 12 Maths Question 11
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)
Solution:
(i) E = {1,3,5}, F = {2,3},E∩F = {3}
P(E) = $frac { 3 }{ 6 }$ ,P(F) = $frac { 2 }{ 6 }$,P(E∩F) = $frac { 1 }{ 6 }$,

Ex 13.1 Class 12 Maths Question 12.
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is girl?
Solution:
Let first and second girls are denoted by G1 and G2 and Boys by B1 and B2.
Sample space
S = {(G1G2),(G1B2),(G2B1),(B1B2)}
Let A = Both the children are girls = {G1G2}
B = youngest child is a girls = {G1G2, B1G2}
C = at least one is a girl = {G1B2, G1G2, B1G2}
A∩B = {G1G2},
A∩C = {G1G2}

Ex 13.1 Class 12 Maths Question 13.
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. 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?
Solution:
The given data may be tabulated as

Ex 13.1 Class 12 Maths Question 14.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:

Let A represents the event “the sum of numbers on the dice is 4”
and B represents the event “the two numbers appearing on throwing two dice are different.”
A = {13,22,31}, n (A) = 3

Ex 13.1 Class 12 Maths Question 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’.
Solution:
Let there be n throws in which a multiple of 3 occurs every time.
Probability of getting a multiple of 3 (i.e. 3 or 6)
in one throw = $frac { 2 }{ 6 }$ = $frac { 1 }{ 3 }$

In each of the following choose the correct answer:

Ex 13.1 Class 12 Maths Question 16.
If P(A) = $frac { 1 }{ 2 }$, P (B) = 0 then P (A | B) is
(a) 0
(b) $frac { 1 }{ 2 }$
(c) not defined
(d) 1
Solution:
P(A) = P(B) = 0
∴ P(A∩B) = 0
$therefore P(A|B)=frac { P(Acap B) }{ P(B) } =frac { 0 }{ 0 }$
Thus option C is correct

Question 17.
If A and B are events such that P(A | B) = P(B | A) then
(a) A⊂B but A≠B
(b) A = B
(c) A∩B = φ
(d) P(A) = P(B)
Solution:
P(A | B) = P(B | A)
Thus option (d) is correct.
$frac { P(Acap B) }{ P(B) } =frac { P(Bcap A) }{ P(A) }$
⇒ P(A) = P(B)

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