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Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.4 | Set 1
• Last Updated : 21 Feb, 2021

### 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 head, B = The last throw result 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.

### 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.

### 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.

### 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.

### 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.

### (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

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