These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1
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) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 }\)
P(F|E) = \(\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }\)
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) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.50 } =\frac { 32 }{ 50 }\) = 0.64
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) = \(\frac { P(A\cap B) }{ P(A) } \Rightarrow 0.4=\frac { P(A\cap B) }{ 0.8 }\)
∴ P(A∩B) = 0.4 x 0.8 = 0.32
(ii) P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.5 } =\frac { 32 }{ 50 } =\frac { 16 }{ 25 }\)
(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98
Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = \(\frac { 5 }{ 13 }\) and P(A|B) = \(\frac { 2 }{ 5 }\).
Solution:
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:
Question 6.
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:
The sample space S = {HHH, HHT, THH, TTH, THH, THT, TTT}
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:
The Sample space S = {HH, TH, HT, TT}
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:
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)}
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:
Let S be the sample space. Then
n(S) = 36
a. E : event of getting a sum greater than 9
E = {(4, 6), (5, 5), (5, 6), (6,4), (6, 5), (6, 6)}
F: event of getting 5 on black die.
F = {(5, 1), (5, 2), (5, 3), (5,4), (5, 5), (5, 6))
E ∩ F = {(5, 5), (5, 6))
\(\mathrm{P}(\mathrm{F})=\frac{6}{36}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}\)
\(P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{6}{36}\right)}=\frac{2}{6}=\frac{1}{3}\)
b. E : event of getting a sum 8
E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F : event of getting a number less than 4 on red die
F= {(1, 1),(2, 1),(3, 1),(4, 1),(5, 1),(6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1,3), (2,3), (3, 3), (4, 3), (5, 3), (6, 3)}
E ∩ F = {(5,3),(6,2)}
\(P(E \cap F)=\frac{2}{36}, P(F)=\frac{18}{36}\)
∴ P(E|F) = \(\frac{P(E ∩ F)}{P(F)}\) = \(\frac{(\frac{2}{36})}{(\frac{18}{36})}\) = \(\frac{1}{9}\)
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:
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 B : The elder boy
G : The elder girl
b : The younger boy
g : The younger girl
The sample space S = {(B, h), (B, g), (G, b),(G, g)}
i. Let E : both children are girls
F : the youngest is a girl
ii. Let E : both children are girls
F : atleast on child is a girl
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:
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 S be the sample space.
∴ n(S) =36
Let E : getting different numbers on the dice
∴ n(E) = 30 ∴P(E) = \(\frac { 30 }{ 36 }\)
Let F : getting the sum of numbers on the dice is 4
∴ F= {(1, 3), (2, 2), (3, 1))
n(F) = 3 ∴ P(F) = \(\frac { 3 }{ 36 }\)
E ∩ F = {(1,3),(3, 1)}
n(E ∩ F) = 2 ∴ P(E ∩ F) = \(\frac{2}{36}\)
\(\mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{\left(\frac{2}{36}\right)}{\left(\frac{30}{36}\right)}=\frac{1}{15}\)
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:
The sample space S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, H). (1, T), (2, H). (2, T), (4, H). (4, T), (5, H), (5, T), (6, 1),(6, 2),(6, 3), (6,4),(6, 5),(6, 6))
Let E: coin shows tail
F : atleast one die shows 3
Now E ∩ F = Φ or P(E ∩ F) = 0
P(E|F) = 0 since P(E ∩ F) = 0
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
∴ P(A|B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0 }{ 0 }\)
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.
\(\frac { P(A\cap B) }{ P(B) } =\frac { P(B\cap A) }{ P(A) }\)
⇒ P(A) = P(B)