Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.4 | Set 2

Question 13. An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3, or 4 on the second toss.

Solution:

According to question:

Given an unbiased die is tossed twice.

Let us consider the events,

A = Getting 4, 5 or 6 on the first toss

B = 1, 2, 3 or 4 on the second toss

⇒ P(A) = 3 / 6 = 1/2

 And, P(B) = 4/6 = 2/3

Now, we have to find that 

P(Getting 4, 5 or 6 on the first toss and 1, 2, 3 or 4 on the second toss)

= P(A ∩ B) = P(A) P(B)

= 1/2 × 2/3 = 2/6 = 1/3

Hence, The required probability = 1/3

Question 14. A bag contains 3 red and 2 black balls. One ball is drawn from it at random. Its color is noted and then it is put back in the bag. A second draw is made and the same procedure is repeated. Find the probability of drawing (i) two red balls

(ii) two black balls,    (iii) first red and second black ball. 

Solution:

According to question:

It is given that,

A bag contain 3 red and 2 black balls.

Let us consider the following events,

A = Getting on red ball.

B = Getting one black ball.

Now, P(A) = 3/5 and, P(B) = 2/5

Now, We have to find that,

(i) P(Getting two red balls)

= P(A) P(A)

= 3/5 × 3/5 = 9/ 25

So, P(Getting two red balls) = 9/25

(ii) P(Getting two black balls)

= P(B) P(B)

= 2/5 × 2/5 = 4/ 25

So, P(Getting two black balls) = 4/25

(iii) P(Getting first red ball and second black ball)

= P(A) P(B)

= 3/5 × 2/5 = 6/ 25

   So, P(Getting first red ball and second black ball) = 6/25

Question 15. Three cards are drawn with replacement from a well-shuffled pack of cards. Find the probability that the cards are drawn are king, queen, and jack.

Solution:

According to question:

Three cards are drawn with replacement. We know that there

are 52 cards in which 4 kings, 4 queens and 4 jacks.

Let us consider the following events

A = Drawing a king

B = Drawing a queen

C = Drawing a jack

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

⇒ P(B) = 4/52 = 1/13

⇒ P(C) = 4/52 = 1/13

Now we have to find that,

P(Cards drawn are king, queen and jack)

Since drawing order are different,

= P(A ∩ B ∩ C) + P(A ∩ C ∩ B) + P(B ∩ A ∩ C) + 

  P(B ∩ C ∩ A) + P(C ∩ A ∩ B) + P(C ∩ B ∩ A)

= P(A) P(B) P(C) + P(A) P(C) P(B) + P(B) P(A) P(C) +

   P(B) P(C) P(A) + P(C) P(A) P(B) +P(C) P(B) P(A) 

Now, putting the values of P(A), P(B) and P(C) in the above expression, then we get

= 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 + 

   1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13

= (1/13 × 1/13 × 1/13) × 6

= 6/2197

Hence, The required probability = 6/2197

Question 16: An artide manufactured by a company consists of two parts X and Y. In the process of manufacture of part X, 9 out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y. Calculate the probability that the assembled product will not be defective.

Solution:

According to question:

It is given that,

Part X has 9 out of 100 defective

So, it is clear that Part X has 91 out of 100 non-defective

Now, Part Y has 5 out of 100 defective

Here, Part Y has 95 out of 100 non-defective

Now let us consider the events,

X = A non-defective part X

Y = A non-defective part Y

So, P(X) = 91/100 and P(Y) = 95/100

We have to find the probability of assembled product will not be defective.

Now, P(Assembled product will not be defective)

 = P(Neither X defective nor Y defective)

= P(X ∩ Y) = P(X) P(Y)

= 91/100 × 95/100

= 0.8645

Hence, The required probability = 0.8645

Question 17. The probability that A hits a target is 1/3 and the probability that B hits it, is 2/5. What is the probability that the target will be hit if each one of A and B shoots at the target?

Solution:

According to question,

Given that,

Probability that A hits a target = 1/3

P(A) = 1/3

And, Probability that B hits a target = 2/5

P(B) = 2/5

We have to find that,

P(Target will be hit)

= 1 – P(target will not be hit)

= 1 – P(Neither A nor B hits the target)

= 1 – P(A‘ ∩ B‘) 

= 1 – P(A‘) P(B‘) = 1 – [1 – P(A)] [1 – P(B)] 

= 1 – [1 – 1/3 ] [1 – 2/5]

= 1 – 2/3 × 3/5 = 1 – 2/5

= 3/5

Hence, The required probability = 3/5

Question 18. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third, and fourth shot are 0.4, 0.3, 0.2, and 0.1 respectively. What is the probability that the gun hits the plane?

Solution:

According to question:

Given that,

An anti-aircraft gun can take a maximum 4 shots at an enemy plane.

Let us consider the following events,

A = Hitting the plane at first shot

B = Hitting the plane at second shot

C = Hitting the plane at third shot

D = Hitting the plane at fourth shot

 ⇒ P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, P(D) = 0.1

We have to find that

P (Gun hits the plane)

= 1 – P(None of four shots hit the plane)

= 1 – P(A‘ ∩ B‘ ∩ C‘ ∩ D‘) 

 = 1 – P(A‘) P(B‘) P(C‘) P(D‘) 

 = 1 – [1 – P(A)] [1 – P(B)] [1 – P(C)] [1 – P(D)]

Now, putting the value of P(A), P(B), P(C) And P(D) in the above expression,

 = 1 – (0.6) (0.7) (0.8) (0.9)

 = 1 – 0.3024

 = 0.6976

Hence, The required probability = 0.6976 

Question 19. The odds against a certain event are 5 to 2 and the odds in favor of another event, independent to the former are 6 to 5. Find the probability that (a) at least one of the events will occur, and (b) none of the events will occur.

Solution:

According to question:

It is given that,

Let us consider the following events, A and B

The odd against a certain event (Say A) are 5 to 2

⇒ P(A‘) = 5/(5 + 2) = 5/7

And, The odd in favor of another event (Say B) are 6 to5

⇒ P(B) = 6/(5 + 6) = 6/11

⇒ P(B‘) =1 – 6/11 = 5/11

Now,

(a) P(At least one of the events will occur)

= 1 – P(None of the events occur)

= 1 – P(A‘∩B‘) = 1 – P(A‘) P(B‘)

= 1 – 5/7 × 5/11 = 1 – 25/77

= 52/77

Hence, the required probability = 52/77

(b) P(None of the events will occur)

= P(A‘ ∩ B‘) = P(A‘) P(B‘)

= 5/7 × 5/11 = 25/77

Hence, the required probability = 25/77

Question 20. A die is thrown thrice. Find the probability of getting an odd number at least once.

Solution:

According to question:

Given that, A die is thrown thrice

Let us consider the following events,

A = Getting an odd number in a throw of die

Since, 1, 3, 5 are odd numbers on die So,

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

P(A‘) = 1/2

We have to find that,

P(Getting an odd number at least once)

= 1 – P(Getting no odd number)

= 1 – P(A‘) P(A‘) P(A‘)

= 1 – 1/2×1/2×1/2 = 1 – 1/8

 = 7/8

Hence, the required probability = 7/8

Question 21. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) Both balls are red

(ii) First ball is black and second is red.

(iii) One of them is black and other is red

Solution:

According to question:

It is given that,

The box contains 10 black balls and 8 red balls.

Then, P(black ball) = 10/18

P(red ball) = 8/18

(i) P(Both balls are red) = 8/18 × 8/18 

= 4/9 × 4/9 = 16/81

(ii) P(First ball is black and second is red.)

= 10/18 × 8/18 = 20/81

(iii) P(One of them is black and other is red)

= 10/18 × 8/18 + 8/18 × 10/18 

= 2 ×(20/81) = 40/81

Question 22. An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting (i) 2 red balls   (ii) 2 blue balls   (iii) one red and one blue ball.

Solution:

According to question:

It is given that,

Urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement.

Let us consider the events,

A = Getting one red ball from the urn.

P(A) = 4/11

B = Getting one blue ball from urn.

P(B) = 7/11

Now, we have to find that,

(i) P(Getting 2 red balls)

= P(A).P(A)

= 4/11×4/11 = 16/121

Hence, the required probability = 16/121

(ii) P(Getting 2 blue balls)

= P(B).P(B)

= 7/11 × 7/11 = 49/121

Hence, the required probability = 49/121

(iii) P(Getting one red and one blue ball)

= P(A).P(B) + P(B).P(A)

= 4/11×7/11 + 7/11× 4/11

= 28/121 + 28/121 = 56/121

Hence, the required probability = 56/121

Question 23. The probabilities of two students A and B coming to the school in time are 3/7 and 5/7 respectively. Assuming that the events, ” A coming in time” and ” B coming in time” are independent, find the probability of only one of them coming to the school on time. Write at least one advantage of coming to school in time.

Solution:

According to question:

It is given that,

The events, ” A coming in time” and ” B coming in time” are independent.

Let us consider the events,

A = A coming in time

B = B coming in time

Now, P(A) = 3/7, and P(A‘) = 1- 3/7 = 4/7

P(B) = 5/7, and P(B‘) = 1- 5/7 = 2/7

Now, we have to find that,

P(Only one coming in time)

⇒ P(A ∩ B‘) + P(A‘ ∩ B) = P(A) × P(B‘) + P(A‘) × P(B)          -(Since, A and B are independent events)

⇒ 3/7 × 2/7 + 4/7 × 5/7 = 6/49 + 20/49 = 26/49

Hence, The required probability = 26/49

The advantages of coming school in time is that you have not to 

give the late fine, you will not miss any part of the lecture and many more.

Question 24: Two dice are thrown together and the total score is noted. The events E, F, and G are ” a total 4″, ” a total of 9 or more”, and ” a total divisible by 5″, respectively. Calculate P(E), P(F), and P(G) and decide which pairs of events, if any, are independent. 

Solution:

According to question:

It is given that,

Two dice are thrown together hence, the total sample space will be 36.

S = {(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),

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

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

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

So, n(S) = 36

Now, let us consider the events,

E be the event of getting a total of 4

E = {(1, 3), (3, 1), (2, 2)}

n(E) = 3

Now, P(E) = n(E)/n(S) = 3/36 = 1/12

Now, F be the event of getting a total 9 or more.

F = {(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}

n(F) = 10

P(F) = n(F)/n(S) = 10/36 = 5/18

Now, G be the event of getting a total divisible by 5.

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

n(G) = 7 

P(G) = n(G) / n(S) = 7/36

Hence, we can see that no pair is independent.

Question 25. Let A and B be two independent events such that P(A) = p1 and P(B) = p2. Describe in words the events whose probabilities are: 

(i) p1p2    (ii) (1 – p1)p2    (iii) 1 – (1 – p1)(1 – p2)    (iv) p1 + p2 = 2p1p2

Solution:

According to question:

The events are said to be independent, if occurrence or non-occurrence of one 

doesn’t affect the probability of the occurrence or non-occurrence of the other.

Now, 

(i) p1p2 = P(A).P(B)

⇒ Both A and B occur.

(ii) (1 – p1)p2 = (1 – P(A)) P(B) = P(A‘) P(B)

⇒ Event A doesn’t occur but B occurs.

(iii) 1 – (1 – p1)(1 – p2) = [1 – (1 – P(A))(1 – P(B))] = (1 – P(A‘) P(B‘))

⇒ At least one of the events A or B occurs.

(iv) p1 + p2 = 2p1p2

⇒ P(A) + P(B) = 2 P(A) P(B)

⇒ P(A) + P(B) – 2 P(A) P(B) = 0

⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0

⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0

⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0

⇒ P(A) (1- P(B)) + P(B) (1 – P(A)) = 0

⇒ P(A) P(B‘) + P(B) P(A‘) = 0

⇒ P(A) P(B‘) + P(B) P(A‘) = 0,   

⇒ P(A) P(B‘) = – P(B) P(A‘)

⇒ Exactly one of A and B occurs.