Class 12 RD Sharma Solutions - Chapter 31 Probability - Exercise 31.4 | Set 1

Question 1(i). A coin is tossed thrice and all the eight outcomes are equally likely. State whether events A and B are independent if, A = The first throw result in the head, B = The last throw results in tail

Solution: 

According to question:

A coin is tossed thrice 

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

Now, 

A = The first throw result in head 

A = {HHT, HTH, HHH, HTT}

Now, B = The last throw result in tail

B = {HHT, HTT, THT, TTT}

A ∩ B = {HHT, HTT}

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

Similarly,

P(B) = 1/2

Now,

P(A ∩ B) = 2/ 8 = 1/ 4

P(A), P(B) = 1/ 2, 1/ 2 

P(A) × P(B) = 1/4

As we know that P(A ∩ B) = P(A)×P(B)

So, A and B are independent events.

Question 1(ii). A coin is tossed thrice and all eight outcomes are equally likely. State whether events A and B are independent if, A = The number of head is odd, B = The number of tails is odd

Solution:

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

A = the number of head is odd, B = the number of tails is odd

So, A = {HTT, THT, TTH, HHH}

B = {THH, HTH, HHT, TTT}

Here, A ∩ B = {} = ∅ 

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

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

And, P(A ∩ B) = 0/8 = 0

Now, P(A) × P(B) = 1/2 × 1/2 = 1/4

So, we can see that 

P(A) × P(B) ≠ P(A ∩ B)

Hence, A and B are not independent events.

Question 1(iii). A coin is tossed thrice and all eight outcomes are equally likely. State whether events A and B are independent if, A = The number of head two, B = The last throw results in head

Solution:

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

Now,

A = The number of head two

A = {HHT, HTH, THH}

Now, B = The last throw results in head

B = {HHH, HTH, THH, TTH}

A ∩ B = {THH, HTH}

P(A) = 3/8 

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

And, P(A ∩ B) = 2/8 = 1/4

Now, P(A) × P(B) = 3/8 ×1/2 = 3/16

So, P(A) × P(B) ≠ P(A ∩ B)

Hence, A and B are not independent events.

Question 2. Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence of 5 on the second die.

Solution:

According to question:

A pair of dice are thrown. So, it has 36 elements in its sample space.

A = Occurrence of number 4 on the first die.

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

B = Occurrence of 5 on the second die.

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

A ∩ B = {(4, 5)}

Now, P(A) = 6/36 = 1/6, P(B) = 6/36 = 1/6 and P(A ∩ B) = 1/36

P(A) × P(B) = 1/6 × 1/6 = 1/36, which is equal to P(A ∩ B).

Hence, A and B are independent events.

Question 3(i). A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. State whether events A and B are independent if, 

A = The card is drawn is a king or queen. 

B = The card drawn is a queen or a jack.

Solution: 

According to question:

A card is drawn form 52 cards. We know that it has 4 Kings, 4 Queen, 4 Jack.

Now, A = The card drawn is a king or queen.

So, P(A) = (4 + 4)/52 = 8/52 = 2/13

B = The card drawn is a queen or a jack.

So, P(B) = (4 + 4)/52 = 8/52 = 2/13

Now, A ∩ B = The drawn card is queen (queen is common in both)

P(A ∩ B) = 4/52 = 1/ 13

Now, P(A) × P(B) = 2/13 × 2/13 = 4/169

We can see that the above expression is not equal to P(A ∩ B). 

So, A and B are not independent events.

Question 3(ii). A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. State whether events A and B are independent if, 

A = The card drawn is black.

B = The card drawn is a king.

Solution:

According to question:

A card is drawn form 52 cards. We know that there are 26 black cards 

in which 2 kings are black.

Now, A = The card drawn is black.

So, P(A) = 26/52 = 1/2

B = The card drawn is a king.

So, P(B) = 4/52 = 1/13

Now, A ∩ B = The drawn card is a black king

P(A ∩ B) = 2/52 = 1/ 26

Now, P(A) × P(B) = 1/2 × 1/13 = 1/26

P(A) × P(B) = P(A ∩ B) 

So, A and B are independent events.

Question 3(iii). A card is drawn from a pack of 52 cards so that each card is equally likely to be selected. State whether events A and B are independent if, 

A = The card is drawn is a spade.

B = The card drawn is an ace.

Solution:

According to question:

A card is drawn form 52 cards. We know that there are 13 spades and 

4 Ace in which 1 card is ace of spade.

Now, A = The card drawn is a spade.

So, P(A) = 13/52 = 1/4

B = The card drawn is an ace.

So, P(B) = 4/52 = 1/13

Now, A ∩ B = The drawn card is an ace of spade.

P(A ∩ B) = 1/52 = 1/ 52

Now, P(A) × P(B) = 1/4 × 1/13 = 1/52

P(A) × P(B) = P(A ∩ B)

So, A and B are independent events.

Question 4: A coin is tossed three times. Let the events A, B, and C be defined as follows:

A = first toss is head, B = second toss is head, C = exactly two heads are tossed in a row.

Check the independence of (i) A and B (ii) B and C (iii) C and A

Solution:

According to question:

A coin is tossed thrice

So, Sample Space = {HTT, HHT, HTH, HHH, THT, THH, TTH, TTT}

A = first toss is head

A = {HHH, HHT, HTH, HTT} 

So, P(A) = 4/8 = 1/2

B = second toss is head

B = {HHH, HHT, THH,THT}

So, P(B) = 4/8 = 1/2

C = exactly two head in a row, i.e., P(C) = {THH, HHT}

So, P(C) = 2/8 = 1/4

Now, A ∩ B = {HHH, HHT}, i.e., P(A ∩ B) = 2/8 = 1/4

Now, B ∩ C = {HHT, THH}, i.e., P(B ∩ C) = 2/8 = 1/4

And, A ∩ C = {HHT}, i.e., P(A ∩ C) = 1/8

(i) P(A) × P(B) = 1/2 × 1/2 = 1/4 

So, P(A) × P(B) = P(A ∩ B)

Hence, A and B are independent events.

(ii) P(B) × P(C) = 1/2 × 1/4 = 1/8

So, P(B)×P(C) ≠ P(B ∩ C)

Hence, B and C are not independent events.

(iii) P(A) × P(C) = 1/2 × 1/4 = 1/8

So, P(A) × P(C) = P(A ∩ C)

Hence, A and C are independent events.

Question 5. If A and B be two events such that P(A) = 1/4, P(B) = 1/3, and P(A ∪ B) = 1/2, Show that A and B are independent events.

Solution:

According to question:

It is given that 

P(A) = 1/4, P(B) = 1/3 and P(A ∪ B) = 1/2

As we know that,

P(A ∩ B) = P(A) + P(B) - P(A ∪ B)

= 1/4 + 1/3 - 1/2

= 1/12

Now, P(A) × P(B) = 1/4 × 1/3 = 1/12

So, P(A) × P(B) = P(A ∩ B)

Hence, A and B are independent events.

Question 6. Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6, Find

(i) P(A ∩ B)      (ii) P(A ∩ B')      (iii) P(A' ∩ B)  

(iv) P(A' ∩ B')      (v) P(A ∪ B) 

(vi) P(A ⁄ B)      (vii) P(B ⁄ A) 

Solution:

According to question:

It is given that A and B are independent events and P(A) = 0.3, P(B) = 0.6

(i) Since, A and B are independents events so,

 P(A ∩ B) = P(A) × P(B)

= 0.3 × 0.6 = 0.18

(ii) P(A ∩ B')= P(A) - P(A ∩ B)

= 0.3 - 0.18 = 0.12

(iii) P(A' ∩ B)= P(B) - P(A ∩ B)

= 0.6 - 0.18 = 0.42

(iv) P(A' ∩ B') = P(A') × P(B')

= [1 - P(A)] × [1 - P(B)]

= [1 - 0.3] × [1- 0.6]

= 0.7 × 0.4 = 0.28

(v) P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

= 0.3 + 0.6 - 0.18 = 0.72

(vi) P(A ⁄ B) = P(A ∩ B) / P(B)

= 0.18/ 0.6 = 0.3

(vii) P(B ⁄ A) = P(A ∩ B) / P(A)

= 0.18/ 0.3 = 0.6

Question 7: If P(not B) = 0.65, P(A ∪ B) = 0.85, and A and B are independent events, then find P(A).

Solution:

According to question:

It is given that P(not B) = 0.65, P(A ∪ B) = 0.85

We know that, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Since, A and B are independent

So, P(A ∩ B) = P(A) × P(B)

Also, P(not B) = 0.65,  

So, P(B) = 0.35            -(Since P(not B) = 1 - P(B))

Hence, We have

⇒ 0.85 = P(A) + 0.35 - P(A) × (0.35)

⇒ 0.5 = P(A)[1-0.35]

⇒ 0.5/ 0.65 = P(A)

So, P(A) = 0.77

Question 8. If A and B are two independent events such that P(A' ∩ B) = 2/15 and P(A ∩ B') = 1/6, then finds P(B).

Solution:

According to question:

It is given that P(A' ∩ B) = 2/15 and P(A ∩ B') = 1/6

Since, A and B are independent,

So, P(A') × P(B) = 2/15 ⇒ [1 - P(A)] P(B) = 2/15         -(1)

and P(A) × P(B') = 1/6 ⇒ P(A) [1 - P(B)] = 1/6         -(2)

From eq(1), We get 

P(B) = 2/15 × 1/(1 - P(A))

On substituting the above value of eq(1) in the eq(2) we get,

P(A)[1 - 2/(15(1-P(A)))] = 1/6

  ⇒ P(A)[15(1 - P(A)) - 2] / [15(1 - P(A))] = 1/6

 ⇒ 6P(A)(13 - 15P(A)) = 15(1 - P(A))

 ⇒ 2P(A)(13 - 15P(A)) = 5 - 5P(A)

 ⇒ 26P(A) - 30 [P(A)]² + 5P(A) - 5 = 0

 ⇒ -30 [P(A)]² + 31 P(A) - 5 = 0         -(3)

This is the form of quadratic equation replace P(A) by x in eq(3)

-30x² + 31x - 5 = 0

30x² - 31x + 5 = 0

So, x = -b ± √b² - 4ac / 2a 

Where, a = 30, b= -31 and c = 5

⇒ x = 31 ± √(-31)² - 4(30)(5) / 60

= 31 ± √961 - 960 / 60

= 30 ± 19 / 60

= 50/60, 12/60 = 5/6, 1/5 

So, P(A) = 5/6 or 1/5

Now, P(A) [1-P(B)] = 1/6

On putting, P(A) = 5/6, we get

5/6[1 - P(B)] = 1/6

⇒ 1 - P(B) = 1/5

⇒ P(B) = 1 - 1/5 = 4/5

Now, putting, P(A) = 1/5, we get

1/5[1 - P(B)] = 1/6

⇒ 1 - P(B) = 5/6

⇒ P(B) = 1 - 5/6 = 1/6

Hence, P(B) = 4/5 or 1/6      

Question 9. A and B are two independent events. The probability that A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Find the probability of occurrence of two events.

Solution:

According to question:

It is given that P(A ∩ B) = 1/6, P(A' ∩ B') = 1/ 3

We know that

P(A' ∩ B') = P(A') × P(B')

1/3 = (1 - P(A))(1 - P(B))

1/3 = 1 - P(B) - P(A) + P(A) P(B)

1/3 = 1 - P(B) - P(A) + P(A ∩ B)

1/3 = 1 - P(B) - P(A) + 1/6

Now, P(A) + P(B) = 1+ 1/6 - 1/3 = 5/6

So, P(A) = 5/6 - P(B)         -(1)

Now, Given that, P(A ∩ B) = 1/6

⇒ P(A) P(B) = 1/6

⇒ [5/6 - P(B)]P(B) = 1/6          -(From eq(1)) 

⇒ 5/6 P(B) - {P(B)}² = 1/6

⇒ {P(B)}² - 5/6 P(B) + 1/6 = 0

⇒ 6 {P(B)}² - 5 P(B) + 1 = 0

After solving this quadratic equation, we get,

⇒ [2P(B) - 1][3P(B) - 1] = 0

⇒ 2 P(B) -1 = 0 or 3P(B) -1 = 0 

⇒ P(B) = 1/2 or P(B) = 1/3

Now,

Using eq(1), 

Putting P(B) = 1/2, ⇒ P(A) = 5/6 - 1/2 = 1/3

Putting P(B) = 1/3, ⇒ P(A) = 5/6 - 1/3 = 1/2

Hence, P(A) = 1/3, P(B) = 1/2 or P(A) = 1/2, P(B) = 1/3.

Question 10. If A and B are two independent events such that P(A ∪ B) = 0.60  and  P(A) = 0.2,  Find P(B).

Solution:

According to question:

It is given that, A and B are independent events and P(A ∪ B) = 0.60, P(A) = 0.2, 

where A and B are independent events. 

So, P(A ∩ B) = P(A) × P(B)

Now, we know that, P(A∪ B) = P(A) + P(B) -P(A∩B)

⇒ 0.6 = 0.2 + P(B) - P(A) × P(B)

⇒ 0.6 - 0.2 = P(B) - 0.2×P(B)          -(Since, P(A) = 0.2)

⇒ 0.4 = 0.8 P(B)

⇒ P(B) = 0.4/0.8 = 0.5

Question 11. A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.

Solution: 

According to question

A dice is tossed twice 

Let us consider the events:

A = Getting a number greater than 3 on first toss

B = Getting a number greater than 3 on second toss

Since, number greater than3 on die are 4, 5, 6

Now, P(A) = 3/6 = 1/2

And, P(B) = 3/6 = 1/2

Now, P(Getting a number greater than 3 on each toss)

= P(A ∩ B) 

= P(A) P(B)

= 1/2 × 1/2 = 1/4

Hence, the required probability = 1/4

Question 12. Given the probability that A can solve a problem is 2/3 and the probability that B can solve the same problem is 3/5. Find the probability that none of the two will be able to solve the problem.

Solution:

According to question,

It is given that, Probability that A can solve a problem = 2/3

⇒ P(A) = 2/3

⇒ P(A') = 1 - 2/3 = 1/3

Now, Probability that B can solve the same problem = 3/5

 ⇒ P(B) = 3/5

⇒ P(B') = 1 - 3/5 = 2/5

Now we have the find the probability that none of them solve the problem

Now, P(None of them solve the problem)

= P(A' ∩ B') = P(A') × P(B')

⇒ 1/3 × 2/5 = 2/15

Hence, The required probability = 2/15

Summary

This chapter introduces the fundamental concepts of probability, including the definitions of sample space, events, and the various rules and principles that govern the calculation of probabilities. It covers topics such as the addition and multiplication principles, conditional probability, Bayes' theorem, and common probability distributions like the binomial and Poisson distributions. The chapter provides a strong foundation for understanding and applying probability in various real-world scenarios, with a focus on problem-solving techniques and the interpretation of results.