These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1/
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)