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# NCERT Solutions class 12 Maths Exercise 13.1

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## NCERT Solutions class 12 Maths Probability

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) =

Ans. (i) P (A  B) = P(A) + P (B) – P (A  B)

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)

=2

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 =

#### 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 =

#### 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) =

Now A  B = {(1, 3), (3, 1)}

= 2

P (A  B) =

Hence, P =

#### 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”.

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

(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}

= 20

P (first die shows a multiple of 3) =

P (first die shows a number which is not a multiple of 3) =

Let A = the coin shows a tail = {(1, T), (2, T), (4, T), (5, T)}

B = at least one die shows a 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}

A  B =

= 4,  = 6,  = 0

P (B) =   and P (A  B) = 0

P =  =

#### In each of the following choose the correct answer:

16. If P (A) =  P (B) = 0, then P is:

(A) 0

(B)

(C) not defined

(D) 1

Ans. P (A) =  P(B) = 0

= 0

P =  =  = not defined

Therefore, option (C) is correct.

#### 17.If A and B are events such that P = P, then:

(A) A

(B) A = B

(C) AB =

(D) P (A) = P (B)

Ans. P = P

P (A) = P (B)

Therefore, option (D) is correct.

## NCERT Solutions class 12 Maths Exercise 13.1

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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### 5 thoughts on “NCERT Solutions class 12 Maths Exercise 13.1”

1. In question no Q8..
the amswer comes out to be 1/6 .Theres a misprint as it is written 1/216.