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# NCERT solutions for class 12 maths exercise 13.1 of chapter 13 Probability

NCERT solutions for class 12 maths exercise 13.1 of chapter 13 Probability are created here for supporting maths students of 12 classes in the preparation of the exams. NCERT solutions for class 12 maths exercise 13.1 of chapter 13 Probability will help the students in clearing their concept of Probability which is one of the important chapters of NCERT maths. Each question of the NCERT solutions for class 12 maths exercises 13.1 of chapter 13 Probability is solved by step by step way so that every student could understand the chapter with the proper understanding that is required in attempting the questions of maths paper in the exam.NCERT solutions is the primary inputs for the preparation of not only the academic exams these are also useful in the preparation of competitive entrance exam.

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### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

## NCERT solutions for class 12 maths exercise 13.1 of chapter 13 Probability

Q1.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).

Ans. We are given that P(E) = 0.6, P(F) = 0.3 and P(E∩F) = 0.2

Let’s find out, probability of event E when event F already occurred, P(E/F)

Applying the formula

Q2. Compute P(A/B) if P(B) =0.5  and P(A∩B)= 0.32

Ans.We are given that P(B) = 0.5, P(A∩B) = 0.32

Let’s  find out P(A/B), probability of event A when event B already occurred

Applying the formula

Q3. 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)

Ans.(i) For getting the value of  P(A∩B) applying the formula of probability

P(A∩B) = 0.4 × 0.8 = 0.32

(iii) Applying the following formula for computing P(A∪B)

P(A∪B) = If P(A) +P(B) – P(A∩B)

=  0.8 +  0.5 – 0.32 = 1.3 -0.32

= 0.98

Q4.Evaluate P(A∪B) if 2 P(A) =P(B)  = 5/13 and P(A/B) = 2/5.

Ans. We are given that P(A/B) = 2/5.

2 P(A) =P(B)  = 5/13 ⇒P(A) = 5/26

Applying the formula of probability

Applying the following formula for computing P(A∪B)

P(A∪B) = If P(A) +P(B) – P(A∩B)

P(A∪B) = 5/26 + 5/13 – 2/13 =5/26 + 3/13 = 11/26

Q5.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)

Ans.

(i) Since we are given P(A) =6/11, P(B) = 5/11, and P(A∪B) =7/11, Applying the following formula for computing P(A∩B)

P(A∪B) =  P(A) +P(B) – P(A∩B)

P(A∩B) =  P(A) +P(B) -P(A∪B)

P(A∩B) = 6/11 +5/11 – 7/11 = 4/11

(ii) Applying the following formula for computing P(A/B)

(iii) Applying the following formula for computing P(B/A)

Q6.Determine P(E/F) in question 6 to 9:

A coin is tossed three times, where

(iii) E: at most two tails F: at least one tail

Ans. If a coin is tossed three times, then sample space of total possible outcomes

S = {HHH,HHT,TTT,TTH,HTH,THH,HTT,THT},n(S) =8

(i) Head on third toss ,E= HHH,TTH,HTH,THH,n(E) =4

Heads on first two tosses,F=HHH,HHT,n(F) =2

We have to find out = P(E/F)

E∩F = HHH,n(E∩F)

P(E) = n(E) /n(S) = 4/8 = 1/2, P(F) = n(F) /n(S) = 2/8 = 1/4,P(E∩F)=n(E∩F)/n(S)=1/8

(ii)  At least two heads,E= HHH,HHT,HTH,THH, P(E) = n(E) /n(S) = 4/8 = 1/2,

At most two heads, F =HHH,HHT,TTT,TTH,HTH,THH,HTT,THT,n(F) =7

P(F)=n(F) /n(S) = 7/8, E∩F= HHH,HHT, HTH,,P(E∩F)=3/8 =3/8

(iii) At most two tails E=HHH,HHT,TTH,HTH,THH,HTT,THT(7),  At least one tail,F= HHT,TTT,TTH,HTH,THH,HTT,THT(7)

E∩F = HHT,TTH,HTH,THH,HTT,THT(6)

P(E) =7/8, P(F) = 7/8,P( E∩F ) = 6/8=3/4

Q7. 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. When two coins are tossed once,then the sample space of outcomes is as following

S ={HH,TH,HT,TT} ie n(S) =4

(i) Tail appears on one coin, E= HT,TH i.e n(E)=2, One coin shows head,F= TH,HT i.e n(F)=2

E∩F = HT,TH i.e n(E∩F)=2

P(E) =n(E) /n(S= 2/4 =1/2, P(F) =n(F) /n(S= 2/4 =1/2, P(E∩F) =n(E∩F) /n(S= 2/4 =1/2

(ii)  No tail appears E= HH,i.e n(E)=1 ,  No head appears,F=TT, i.e n(F)=1

E∩F = Φ i.e n(E∩F)= 0

P(E) =n(E) /n(S)= 1/4 , P(F) =n(F) /n(S)= 1/4 , P(E∩F) =n(E∩F) /n(S= 0/4 =0

Q8. A die is thrown three times.

(i) E: 4 appears on the third toss

(ii) F: 6 and 5 appears respectively on the first two tosses

Ans. When a die is thrown 3 times , then the sample space of outcomes is =6³=216,n(S) =216

(i) 4 appears on the third toss, E = {114,124,134,144,154,164,214,224,234,244,254,264,314,324,334,344,354,364,414,424,434,444,454,464,514,524,534,544,554,564,614,624,634,644,654,664} i.e n(E) =36

6 and 5 appears respectively on the first two tosses, F= {651,652,653,654,655,656}

n(F) = 6 and  E∩F = 654, n(E∩F) = 1

P(E) = n(E)/n(S) = 36/216 =1/6, P(F) = n(F)/n(S) = 6/216 =1/36, P(E∩F) =  n(E∩F)/n(S) = 1/216

Q9. Mother, father and Son line up at random for a family picture:

E: son on one end, F: father in the middle

Ans. The sample space of outcomes arranging father,mother and son =3! = 6,n(S)=6

Mother(m), father (f) (and Son(s) line at random ,son on one end,E ={sfm,smf,fms,mfs}, n(E) =4

Father in the middle,F =sfm,mfs, n(F) = 2

E∩F = sfm,mfs,n(E∩F) =2

P(E) = n(E)/n(S) = 4/6 =2/3, P(F) = n(F)/n(S) = 2/6 =1/3, P(E∩F) =  n(E∩F)/n(S) = 2/6=1/3

Q10. A black and a red die rolled.

(i) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(ii) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Ans. When two dice are thrown then the outcomes of sample space = 6²= 36,n(S)=36

(i) Let E is the event of  obtaining a sum greater than 9, then E={(4,6),(5,5),(5,6),(6,6),(6,5),(6,4)}

Let F is the event of black die resulted in a 5, F ={(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)},n(F) =6

E∩F =  (5,5),(5,6), n(E∩F) = 2

P(F) = n(F)/n(S) = 6/36 =1/6,  P(E∩F) =  n(E∩F)/n(S) = 2/36=1/18

(ii)Let E  is the event of obtaining the sum 8,E={(2,6),(3,5),(4,4),(5,3),(6,2), n(E) =5

Let F is the event of red die resulted in a number less than 4, F ={(1,1),(1,2),(1,3)(2,1),(2,2),(2,3)(3,1),(3,2),(3,3),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3), n(F) = 18

E∩F =  (5,3),(6,2), n(E∩F) = 2

P(F) = n(F)/n(S) = 18/36 =1/2,  P(E∩F) =  n(E∩F)/n(S) = 2/36=1/18

You can try following questions

Q11. 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) 22and P(G|E)

(iii) P((E ∪ F) |G) and P(E ∩ F)|G)

Q12. 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?

Q13. An instructor has a question bank consisting of 300 easy True/false multiple choice questions, 200 difficult True/False questions, 500 easy multiple-choice questions, and 400 difficult 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 question?

Q14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of the numbers on the dice is 4’.

Q15. 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’.

Q16. If P(A) = 1/2, P(B) = 0 then P(A | B) is

(i) 0

(ii) 1/2

(iii) not defined

(iv) 1

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 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

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