Open in App
Not now

# Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.3 | Set 1

• Last Updated : 03 Mar, 2021

### Question 1. If P(A) = 7 /13, P(B) = 9/13 and P(A âˆ© B) = 4/13, find P(A/B).

Solution:

Given: P(A) = 7/13, P(B) = 9/13, and P(Aâˆ© B) = 4/13

We know that, P(A/B) = P(A âˆ© B)/P(B)

= (4/13) Ã· (9/13)

= 4/9

### Question 2. If A and B are events such that P(A) = 0.6, P(B) = 0.3, and P(A âˆ© B) = 0.2, find P(A/B) and P(B/A).

Solution:

Given: P(A) = 0.6, P(B) = 0.3, P(A âˆ© B) = 0.2

We know that P(A/B) = P(A âˆ© B)/P(B)

= 0.2/0.3

= 2/3

and P(B/A) = P(A âˆ© B)/P(A)

= 0.2/0.6

= 1/3

### Question 3. If A and B are two events such that P(A âˆ© B) = 0.32 and P(B) = 0.5, find P(A/B).

Solution:

Given: P(A âˆ© B) = 0.32 and P(B) = 0.5

We know that P(A/B) = P(A âˆ© B)/P(B)

= 0.32/0.5

= 0.64

### Question 4. If P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6, find P(A/B) and P(A âˆª B).

Solution:

Given: P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6

We know that, P(B/A) = P(A âˆ© B)/P(A)

P(A âˆ© B) = P(B/A) Ã— P(A)

P(A âˆ© B) = 0.6 Ã— 0.4

P(A âˆ© B) = 0.24

Therefore, P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 0.4 + 0.8 – 0.24

= 0.96

And, P(A/B) = P(A âˆ© B)/P(B)

= 0.24/0.8

= 0.3

### (i) P(A) = 1/3, P(B) = 1/4, and P(A âˆª B) = 5/12, find P(A/B) and P(B/A).

Solution:

We know that, P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

P(A âˆ© B) = P(A) + P(B) – P(A âˆª B)

P(A âˆ© B) = 1/3 + 1/4 – 5/12 = 2/12

Therefore, P(A/B) = P(A âˆ© B)/P(B) = (2/12) Ã· (1/4) = 2/3

and, P(B/A) = P(A âˆ© B)/P(A) = (2/12) Ã· (1/3) = 1/2

### (ii) P(A) = 6/11, P(B) = 5/11 and P(A âˆª B) = 7/11, find P(A âˆ© B), P(A/B) and P(B/A).

Solution:

We know that, P(Aâˆª B) = P(A) + P(B) – P(A âˆ© B)

P(A âˆ© B)=P(A) + P(B) – P(A âˆª B)

P(A âˆ© B) = 6/11 + 5/11 – 7/11

= 4/11

Therefore, P(A/B) = P(A âˆ© B)/P(B)

= (4/11) Ã· (5/11) = 4/5

and, P(B/A) = P(A âˆ© B)/P(A)

= (4/11) Ã· (6/11) = 2/3

### (iii) P(A) = 7/13, P(B) = 9/13 and P(A âˆ© B) = 4/13, find P(A’/B).

Solution:

We know that, P(A’ âˆ© B) = P(B) – P(A âˆ© B)

= 9/13 – 4/13

= 5/13

Therefore, P(A’/B) = P(A’ âˆ© B)/P(B)

= (5/13) Ã· (9/13)

= 5/9

### (iv) P(A) = 1/2, P(B) = 1/3 and  P(A âˆ© B) = 1/4, find P(A/B), P(B/A), P(A’/B), and P(A’/B’).

Solution:

P(A/B) = P(A âˆ© B)/P(B)

= (1/4) Ã· (1/3)

= 3/4

P(B/A) = P(A âˆ© B)/P(A)

= (1/4) Ã· (1/2)

= 1/2

P(A’/B) = (P(B) – P(A âˆ© B))/P(B)

= (1/3 – 1/4) Ã· (1/3)

= 1/4

P(A’/B’) = P(A’ âˆ© B’)/P(B’)

= (1 – P(A’ âˆª B’))/(P(A) – P(A âˆ© B))

= (1 – P(A) – P(B) + P(A âˆ© B))/(P(A) – P(A âˆ© B))

= (1 – 1/2 – 1/3 + 1/4)/(1/2 – 1/4)

= 5/4

### Question 6. If A and B are two events such that 2 Ã— P(A) = P(B) = 5/13 and P(A/B) = 2/5, find P(A âˆª B).

Solution:

Given, 2 Ã— P(A) = P(B) = 5/13

P(A) = 5/26

P(A/B) = P(A âˆ© B)/P(B)

2/5 = P(A âˆ© B) Ã· (5/13)

P(A âˆ© B) = (2/5) Ã· (5/13)

P(A âˆ© B) = 2/13

We know that,

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 5/26 + 5/13 – 2/13

= 11/26

### (i) P(A âˆ© B)

Solution:

We know that, P(Aâˆª B) = P(A) + P(B) – P(A âˆ© B)

P(A âˆ© B) = P(A) + P(B) – P(A âˆª B)

= 6/11 + 5/11 – 7/11

= 4/11

### (ii) P(A/B)

Solution:

P(A/B) = P(A âˆ© B)/P(B)

= (4/11) Ã· (5/11)

= 4/5

### (iii) P(B/A)

Solution:

P(B/A) = P(A âˆ© B)/P(A)

= (4/11) Ã· (6/11)

= 2/3

### (iii) A = At most two tails, B = At least one tail.

Solution:

Sample space of three coins is given by

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

(i) A = Heads on third toss = {HHH, HTH, THH, TTH}

B = Heads on first two tosses = {HHH, HHT}

P(A âˆ© B) = {HHH}

âˆ´ P(A/B) = P(A âˆ© B)/P(B) = 1/2

(ii) A = At least two heads = {HHH, HTH, THH, HHT}

B = At most two heads = {HHT, HTH, TTH, HHT, HTT, THT, TTT}

P(A âˆ© B) = {HTH, THH, HHT}

âˆ´ P(A/B) = P(A âˆ© B)/P(B) = 3/7

(iii) A = At most two tails = {HHH, HTH, THH, TTH, HHT, HTT, THT}

B = At least one tail  = {HTH, THH, TTH, HHT, HTT, THT, TTT}

P(A âˆ© B)  = {HTH, THH, TTH, HHT, HTT, THT}

âˆ´  P(A/B) = P(A âˆ© B)/P(B) = 6/7

### (ii) A = No tail appears, B = No head appears

Solution:

Sample space of two coins is given by

{HH, HT, TH, TT}

(i) A = Tail appears on one coin = {HT, TH}

B = One coin shows head = {HH, HT, TH}

P(A âˆ© B)  = {HT, TH}

âˆ´ P(A/B) = P(A âˆ© B)/P(B)

= 2/2

= 1

(ii) A = No tail appears = {HH}

B = No head appears= {TT}

P(A âˆ© B)  = { }

âˆ´ P(A/B) = P(A âˆ© B)/P(B) = 0

### A = 4 appears on the third toss, B = 6, and 5 appear respectively on the first two tosses.

Solution:

A = 4 appears on the third toss = { (1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),

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

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

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

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

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

B = 6 and 5 appear respectively on first two tosses

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

(A âˆ© B) = {(6, 5, 4)}

âˆ´ P(A/B) = P(A âˆ© B)/P(B) = 1/6

âˆ´ P(B/A) = P(A âˆ© B)/P(A) = 1/36

### Question 11. Mother, father, and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P(A/B) and P(B/A).

Solution:

Let Mother = M, Father = F, and son = S

Sample Space = {FMS, FSM, MFS, MSF, SFM, SMF}

A = Son on one end = {FMS, MFS, SFM, SMF}

B = Father in the middle = {MFS, SFM}

P(A âˆ© B) = {MFS, SFM}

âˆ´ P(A/B) = P(A âˆ© B)/P(B) = 2/2 = 1

âˆ´ P(B/A) = P(A âˆ© B)/P(A) = 2/4 = 1/2

### Question 12. A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Solution:

Let the sample space of the experiment is {(1, 1), (1, 2), (1, 3), . . . .,(6, 6)} consisting of 36 outcomes

P(A) =  P(Sum = 6) = 5/36

P(B) = P(4 has appeared at least once) = 11/36

P(B/A) = P(A âˆ© B)/P(A) = (2/36) Ã· (5/36) = 2/5

### Question 13. Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Solution:

Two dice are thrown.

A = Sum on the dice is 8

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

B = Second die always exhibits 4

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

P(A âˆ© B) = {(4, 4)}

P(A/B) = P(A âˆ© B)/P(B)

= 1/6

### Question 14. A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Solution:

A = Sum on two dice equals 7 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

B = Second die always exhibits an odd number = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)

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

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

(A âˆ© B) = {(6, 1), (2, 5), (4, 3)}

P(A/B) = P(A âˆ© B)/P(B)

= 3/18

= 1/6

My Personal Notes arrow_drop_up
Related Articles
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}