Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Exercise 13.1

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1 provides solutions for all the questions in the NCERT textbook with simple and easy to understand solution. Our comprehensive solutions will guide you step-by-step through each question in Exercise 13.1, making complex concepts easy to grasp. Whether you're preparing for exams or just looking to strengthen your math skills, these solutions are here to assist you every step of the way.

Chapter 13 – Probability Exercise 13.1

Question 1: Given that E and F are 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

So, P(F ∩ E) = 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 = 2/6 = 1/3

Question 2: Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

Solution:

Now, P(A|B) = (P(A ∩ B))/(P(B)) = (0.32)/(0.5) = 32/50 = 16/25

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:

Given: P(A) = 0.8 ,

P(B) = 0.5 and

P(B|A) = 0.4

(i) Now, P(B|A) = (P(B ∩ A))/(P(A))

⇒ 0.4 = (P(B ∩ A))/(0.8)

⇒ P(A ∩ B) = 0.4 × 0.8 P(A ∩ B) = 0.32

(ii) P(A|B) = (?(? ∩ ?))/(?(?)) = (0.32)/(0.5) = 32/50 = 0.64

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.8 + 0.5 – 0.32 = 1.3 – 0.32 = 0.98

Question 4: Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5

Solution:

Given, 2P(A) = P(B) = 5/13 Now, 2 P(A) = 5/13

∴ P(A) = 1/2 × 5/13 = ?/?? And, P(B) = 5/13

Now, P(A|B) = (?(? ∩ ?))/(?(?)) ?/? = (?(? ∩ ?))/(?/??) "P(A ∩ B)" = 2/5 × 5/13 "P(A ∩ B)" = ?/??

Also, "P(A"∪"B)" = P(A) + P(B) – "P(A ∩ B)" = 5/26 + 5/13 – 2/( 13) = 5/26 + 3/13 = 5/26 + 6/26 = ??/?? ∴ "P(A "∪" B)" = ??/??.

Question 5: If P(A) = 6/11 , P(B) = 5/11 and P(A ∪ B) 7/11 , Find

(i) P(A ∩ B)

(ii) P(A|B)

(iii) P(B|A)

Solution:

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

Now, (i) P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

7/11 = 6/11 + 5/11 - P(A ∩ B)

P(A ∩ B) = 11/11 - 7/11

P(A ∩ B) = 4/11

(ii) P(A|B) = (P(A ∩ B))/(P(B)) = (4/11)/(5/11) = 4/5

(iii) P(B|A) = (P(B ∩ A))/(P(A)) = (4/11)/(6/11) = 2/3

Determine P(E|F) in Excercise 6 to 9.

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 Coin is tossed three times

(iii) E : at most two tails, F : at least one tail Coin is tossed three times .

Coin is tossed three times, sample space is

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

(i)

E: head on third toss

E = {HHH, HTH, THH, TTH}

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

F: heads on first two tosses

F = {HHH, HHT}

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

Also, E ∩ F = {HHH}

P(E ∩ F) = 1/8

We need to find P(E|F)

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

(ii)

E: at least two heads

E = {HHT, THH, HTH, HHH}

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

F: at most two heads

F = {HHT, THH, HTH, TTH, THT, HTT, TTT}

P(F) = 7/8

Also, E ∩ F = {HHT, THH, HTH}

P(E ∩ F) = 3/8

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

(iii)

E: at most two tails

E = {HHH, HHT, HTH, THH, TTH, THT, HTT}

P(E) = 7/8

F: at least one tail

F = {HHT, HTH, THH, TTH, THT, HTT, TTT}

P(F) = 7/8

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

P(E ∩ F) = 6/8

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

Question 7: Two coins are tossed once, where

(i) E : tail appears on one coin, F : one coin shows headTwo coins are tossed once

S = {(H, H), (H, T), (T, H), (T, T)}

(i)

We need to find the probability that tail appears on one coin given that one coin shows head

Let E: tail appears on one coin

F: one coin shows head

We need to find P(E|F)

Event E

E = {(H, T), (T, H)}

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

Event F

F = {(H, T), (T, H)}

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

Also, E ∩ F = {(H, T), (T, H)}

P(E ∩ F) = 2/4 = 1/2

Now, P(E|F) = (P(E ∩ F))/(P(F)) = (1/2)/(1/2) = 1

(ii)

We need to find the probability that no tail appears, if known that no head appears

Let E: no tail appears

F: no head appears

We need to find P(E|F)

E = {(H, H)}

P(E) = 1/4

F = {(T, T)}

P(F) = 1/4

Also, E ∩ F = ϕ

∴ P(E ∩ F) = 0

Now, P(E|F) = (P(E ∩ F))/(P(F)) = 0/(1/4) = 0

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

A die is thrown 3 times

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

Total cases = 6 × 6 × 6 = 216

Given,

A: 4 on the third throw

B: 6 on the first & 5 on the second throw

We need to find the probability of A, given that B has already occurred i.e. P(A|B)

Finding P(B)

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

P(B) = 6/216

Finding P(A ∩ B)

A ∩ B = {(6, 5, 4)}

P(A ∩ B) = 1/216

Now, P(A|B) = (P(A ∩ B))/(P(B)) = (1/216)/(6/216) = 1/6

Question 9: Mother, father and son line up at random for a family picture

E : son on one end,

F : father in middle

We need to find the probability that the son is on one end, given that the father is in the middle.

So, let E: Son on one end

F: Father in middle

We need to find P(E|F)

Event E

E = {(M,F,S), (F,M,S), (S,M,F), (S,F,M)}

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

Event F

F = {(M, F, S), (S, F, M)}

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

Also, E ∩ F = {(M,F,S), (S,F,M)}

∴ P(E ∩ F) = 2/6 = 1/3

Now, P(E|F) = (P(E ∩ F))/(P(F)) = (1/3)/(1/3) = 1

Question 10: A black and a red dice 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)

We need to find the Probability of obtaining a sum greater than 9, given that the black die resulted in 5

Let E: Sum of numbers greater than 9

F: 5 appeared on the black die

We need to find P(E|F)

Event E

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

P(E) = 6/36

Event F

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

P(F) = 6/36

Also, E ∩ F = {(5, 5), (5, 6)}

So, P(E ∩ F) = 2/36

Now, P(E|F) = (P(E ∩ F))/(P(F)) = (2/36)/(6/36) = 1/3

∴ Required probability is 1/3

(b)

Let E: Sum of numbers is 8

F: Number on red die is less than 4

We need to find P(E|F)

Event E

E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

P(E) = 5/36

Event F

F = {(1, 1), (2, 1), (3, 1), ..., (6, 1), (1, 2), (2, 2), ..., (6, 2), (1, 3), (2, 3), ..., (6, 3)}

P(F) = 18/36

Also, E ∩ F = {(5, 3), (6, 2)}

So, P(E ∩ F) = 2/36

Now, P(E|F) = (P(E ∩ F))/(P(F))

= (2/36)/(18/36)

= 1/9

Therefore, the required probability is 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:

(i)

We need to find P(E|F) and P(F|E)

Given: E ∩ F = {3}

P(E ∩ F) = 1/6

P(E|F) = (P(E ∩ F))/(P(F))

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

= 1/2

P(F|E) = (P(F ∩ E))/(P(E))

= (P(E ∩ F))/(P(E)) (By commutative property)

= (1/6)/(1/2) (Since P(E) = 1/2)

= 1/3

(ii)

P(E)= ∣E∣/6 ​= 3/6 ​= 1/2

P(G)= ∣G∣/6 ​= 4/6 ​= 2/3

G∩E = E∩G = {3, 5}

⇒ P(E∩G) = P(G∩E) = 2/6 = 1/3

Thus, P(E∣G) = P(E∩G)/P(G)​ = [1/3]/[2/3] = 1/2

Thus, P(G|E) = P(G∩E)/P(E)​ = [1/3]/[1/2] = 2/3

(iii)

Let E ∪ F = A

So, A = {1, 2, 3, 5}

P(A) = 4/6

Now, A ∩ G = {2, 3, 5}

So, P(A ∩ G) = 3/6

Thus, P(A|G) = (P(A ∩ G))/(P(G)) = (3/6)/(4/6) = 3/4

Therefore, P((E ∪ F) | G) = 3/4

Let E ∩ F = B

So, B = {3}

P(B) = 1/6

Now, B ∩ G = {3}

So, P(B ∩ G) = 1/6

Thus, P(B|G) = (P(B ∩ G))/(P(G)) = (1/6)/(4/6) = 1/4

Therefore, P((E ∩ F) | G) = 1/4

Question 12: Assume that each born child 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 a girl?

Solution:

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

We need to find the probability that the children are girls, given that at least one is a girl.

Let E: both the children are girls

F: at least one child is a girl

E = {(g, g)}

P(E) = 1/4

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

P(F) = 3/4

Also, E ∩ F = {(g, g)}

So, P(E ∩ F) = 1/4

Now, P(E|F) = (P(E ∩ F))/(P(F)) = (1/4)/(3/4) = 1/3

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:

Let Easy questions be denoted by E, Difficult questions by D, MCQs by M & True/False questions by T

Given:

E ∩ T = 300, D ∩ T = 200, E ∩ M = 500, D ∩ M = 400

We need to find the probability that it will be an easy question, given that it is a MCQ, i.e., P(E|M)

P(E ∩ M) = Probability that question is Easy MCQ = (Total Easy MCQ Questions)/(Total questions) = 500/1400 = 5/14

And, P(M) = (Total MCQ questions)/(Total questions) = (500 + 400)/1400 = 900/1400 = 9/14

Thus, P(E|M) = (P(E ∩ M))/(P(M)) = (5/14)/(9/14) = (5/14) × (14/9) = 5/9

Therefore, the required probability is 5/9.

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 :

Two Dice are thrown We need to find the Probability that the sum of numbers on the dice is 4, given that the two numbers are different.

Let E: Sum of numbers is 4

F: Numbers are different

We need to find P(E|F)

Also, E ∩ F = {(1, 3), (3, 1)}

P(E ∩ F) = 2/36 = 1/18

Event E

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

P(E) = 3/36 = 1/12

Event F

P(F) = 30/36 = 5/6 (There are 30 outcomes with different numbers on the two dice out of 36 total outcomes)

Now, P(E|F) = (P(E ∩ F))/(P(F)) = (1/18)/(5/6) = 2/30 = 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:

A die is thrown. If multiple of 3 comes up, the die is thrown again. If any other number comes up, a coin is tossed.

Let E: Coin shows a tail

F: At least one die shows 3

We need to find P(E|F)

Event E

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

P(E) = 1/12 + 1/12 + 1/12 + 1/12 = 4/12 = 1/3

Event F

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

P(F) = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 7/36

Also, E ∩ F = ϕ

So, P(E ∩ F) = 0

Now, P(E|F) = P(E ∩ F)/(P(F)) = 0/(7/36) = 0

Your reasoning and calculations are correct. Well done!

Question 16: If P (A) = 1/2 , P(B) = 0, then P (A|B) is:

(A) 0 (B) 1/2 (C) not defined (D) 1

Solution:

If P(B) = 0, then B = ϕ (the null set). This implies A ∩ B = ϕ and P(A ∩ B) = 0.

Then, P(A|B) = (P(A ∩ B))/(P(B)) = 0/0, which is undefined.

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)

⇒ (P(A ∩ B))/(P(B)) = (P(B ∩ A))/(P(A))

⇒ (P(A ∩ B))/P(B) = (P(A ∩ B))/P(A)

⇒ 1/P(B) = 1/P(A)

⇒ P(B) = P(A)

Therefore, if P(A|B) = P(B|A), it means P(A) = P(B). This proves that the correct option is (D).

Summary

Chapter 13 on Probability in NCERT Class 12 Mathematics Part II covers fundamental concepts of probability theory, including sample spaces, events, and probability calculations. It explores various probability scenarios, from simple experiments like coin tosses and die rolls to more complex situations involving multiple events and conditional probabilities. The chapter emphasizes the practical application of probability in real-world situations and provides a foundation for more advanced statistical concepts.

• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 13 – Probability Exercise 13.3
• Class 12NCERT Solutions - Mathematics Part ii – Chapter 13 – Probability Exercise 13.2