NCERT Solutions for Class 12 Maths Exercise 13.1 Chapter 13 Probability – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1 (Ex 13.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 13 Probability Exercise 13.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.1) Exercise 13.1

1. Given that E and F are events such that P (E) = 0.6, P (F) = 0.3 and P (EF) = 0.02, find P (E) and P.

Ans. Given: P (E) = 0.6, P (F) = 0.3 and P (E

F) = 0.2

P = =

And P = =

2. Compute P if P (B) = 0.5 and P (AB) = 0.32

Ans. Given: P (B) = 0.5, P (A

B) = 0.32

P = = = 0.64

3. If P (A) = 0.8, P (B) = 0.5 and P = 0.4, find:

(i) P (AB)

(ii) P

(iii) P (AB)

Ans. (i) P (A

B) = P

. P (A) = 0.4 x 0.8 = 0.32

(ii) P = = = 0.64

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

4. Evaluate P (AB) if 2P (A) = P (B) = and P =

Ans. Given: 2 P (A) = P (B) =

, P

P (A) =

Now P (A B) = P . P (B) =

And P (A B) = P(A) + P (B) – P (A B) = =

5. If P (A) = P (B) = and P (AB) = Find (i) $P (A \cap B)$ , (ii) $P (A | B)$ , (iii) $P (B | A)$

Ans. (i) P (A

B) = P(A) + P (B) – P (A

P (A B)

P (A B) =

(ii) P = =

(iii) P = =

6. Determine P : A coin is tossed three times.

(i) E : heads 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.

Ans. A coin tossed three times, i.e.,

S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)

= 8

(i) E : heads on third toss

E = (TTH, HTH, THH, HHH)

P (E) =

F : heads on first two tosses

F = (HHT, HHH)

P (F) =

E F = (HHH)

(E F) = 1

P (E F) =

And P =

(ii) E : at least two heads

E = (HHT, HTH, THH, HHH)

P (E) =

F : at most two heads

F = (TTT, HTT, THT, TTH, HHT, HTH, THH)

= 7

P (F) =

E F = (HHT, HTH, THH)

(E F) = 3

P (E F) =

And P =

(iii) E : at most two tails

E = (HTT, THT, TTH, HHT, HTH, THH, HHH)

P (E) =

F : at least one tail

F = (TTT, HTT, THT, TTH, HHT, HTH, THH)

= 7

P (F) =

E F = (HTT, THT, TTH, HHT, HTH, THH)

(E F) = 6

P (E F) =

And P =

7. Determine P : 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.

Ans. S = (HH, TH, HT, TT)

= 4

(i) E : tail appears on one coin

E = (TH, HT)

P (E) =

F : one coin shows head

F = (TH, HT) = 2

P (F) =

E F = (TH, HT) (E F) = 2

P (E F) =

And P =

(ii) E : no tail appears

E = (HH)

P (E) =

F : no head appears

F = (TT) = 1

P (F) =

E F = (E F) = 0

P (E F) =

And P = $\frac{P (E \cap F)}{P (F)}$ = $\frac{0}{1 / 4} = 0$

8. Determine P : A dice is thrown three times.

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

Ans. Since a dice has six faces. Therefore

= 6 x 6 x 6 = 216

E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)

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

= 1 x 1 x 6 = 6

P (F) =

E F = (6, 5, 4)

(E F) = 1

P (E F) =

And P = $\frac{P (E \cap F)}{P (F)}$ = $\frac{1 / 216}{6 / 216} = \frac{1}{6}$

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

E : Son on one end, F : Father in middle.

Ans. S = (MFS, MSF, SFM, SMF, FMS, FSM)

E : Son on one end

E = (MFS, SFM, SMF, FMS)

F : Father in middle

F = (MFS, SFM) = 2

P (F) =

E F = (MFS, SFM) (E F) = 2

P (E F) =

And P =

10. A black 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.

Ans. (a)

= 6 x 6 = 36

Let A represents 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) = 10

P (A) =

B = (51, 52, 53, 54, 55, 56) = 6

P (B) =

A B = (55, 56) = 2

P (A B) =

P =

(b) Let A denoted the sum is 8

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

B = Red die results in a number less than 4, either first or second die is red

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

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

P (B) =

A B = {(2, 6), (3, 5) = 2

P (A B) = =

P =

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

(i) P and P

(ii) P and P

(iii) P and P

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

= 6

E = (1, 3, 5) F = (2, 3) G = (2, 3, 4, 5)

= 3 = 2 = 4

(i) P (E) = P (F) =

E F = (3) = 1

P (E F) =

P = and P =

(ii) P (E) = P (G) =

E G = (3, 5) = 2

P (E G) =

P = and P =

(iii) P (G) =

E F = (1, 2, 3, 5) and G (2, 3, 4, 5)

(E F) G = (2, 3, 5) = 3

P =

Again E F = (3)

(E F) G = (3) = 1

P =

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 a girl?

Ans. Let first and second girl are denoted by G₁ and G₂ and boys B₁ and B₂.

S = {(G₁G₂), (G₁B₂), (G₂B₁), (B₁B₂)}

Let A = Both the children are girls = (G₁G₂)

B = Youngest child is girl = {(G₁G₂), (B₁G₂)}

C = at least one is a girl = {(G₁B₂), (G₁G₂), (B₁G₂)}

A B = (G₁G₂) P (A B) =

A C = (G₁G₂) P (A C) =

P (B) = and P (C) =

(i) P =

(ii) P =

13. An instructor has a test 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 test bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Ans. Number of easy True/False questions = 300

Number of difficult True/False questions = 200

Number of easy multiple choice questions = 500

Number of difficult multiple choice questions = 400

Total number of all such questions = = 1400

Let E represents an easy question and F represents a multiple choice question.

= 300 + 500 = 800 and = 500 + 400 = 900

P (F) =

= 500 P (E F) =

P =

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’.

Ans. S = (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)

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

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

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

= 36

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”.

Therefore, A = {(1, 3), (2, 2), (3, 1)} = 3

P (A) =

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

= 30

P (B) =