# Class 12 RD Sharma Solutions – Chapter 31 Probability – Exercise 31.2

• Last Updated : 13 Jan, 2022

### Question 1: From a pack of 52 cards, Two are drawn one by one without replacement. Find the probability that both of them are kings.

Solution:

Let the desired events be,

A = first card is king

B = second card is also king

and probability of the event A is, P(A).

P(A) = 4/52 = 1/13 [There are 4 kings in the 52 cards set]

As the king is withdrawn without replacement there are three kings left only.

And the number of cards left is 51 as well.

Hence,

P(B/A) = 3/51 = 1/17

Hence, the required probability,

P(A∩B) = P(A) × P(B/A) = (1/13) × (1/17)

or, P(A∩B) = 1/221 (ans)

### Question 2: From a pack of 52 cards, four are drawn one by one without replacement. Find the probability that all of them are Aces.

Solution:

Let the desired events be,

A = an ace in the first draw

B = an ace in the second draw

C = an ace in the third draw

D = an ace in the fourth draw

The probability of the event x is P(x)

P(A) = 4/52 = 1/13 [There are 4 aces and 52 cards remaining]

P(B/A) = 3/51 = 1/17 [There are 3 aces and 51 cards remaining]

P(C/A∩B) = 2/50 = 1/25 [There are 2 aces and 50 cards remaining]

P(D/A∩B∩C) = 1/49 [There is 1 ace and 49 cards remaining]

The required probability,

P(A∩B∩C∩D) = P(A) × P(B/A) × P(C/A∩B) × P(D/A∩B∩C)

=(1/13) × (1/17) × (1/25) × (1/49)

=1/270725 (ans)

### Question 3: Find the chance of drawing 2 white balls in succession from a bag containing 5 red and 7 white balls. The ball first-drawn never get replaced.

Solution:

Let the desired events be,

A = White ball at first draw

B = White ball at second draw

The probability of the event x is P(x).

P(A) = 7/12 [Any of the seven white balls from the 12-ball-set]

P(B/A) = 6/11 [6 white balls remaining and 11 balls left]

The required probability,

P(A∩B) = P(A) × P(B/A)

= (7/12) × (6/11) = 7/22 (ans)

### Question 4: A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.

Solution:

Let the desired events be,

A = Even number at first draw

B = Even number at second draw

The probability of the event x is P(x).

P(A) = 12/25 [Any of the12 even numbers less than 25]

P(B/A) = 11/24 [Any of the 11 remaining even number tickets out of 24]

The required probability,

P(A∩B) = P(A) × P(B/A)

= (12/25) × (11/24) = 11/50 (ans)

### Question 5: From a deck of cards, three cards are drawn one by one without replacement. Find the probability that it’s a card of aspade.

Solution:

Let the desired events be,

A = a spade in the first draw

B = a spade in the second draw.

C = a spade in the third draw.

P(A) = 13/52 = 1/4 [There are 13 spades and 52 cards remaining]

P(B/A) = 12/51 = 4/17 [There are 12 spades and 51 cards remaining]

P(C/A∩B) = 11/50 [There are 11 spades and 50 cards remaining]

The required probability,

P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)

=(1/4) × (4/17) × (11/50) = 11/850 (ans)

### (i) Two cards are drawn without replacement from a pack of 52 cards. Find the probability that both are kings.

Solution:

Let the desired events be,

A = first card is king

B = second card is also king

and probability of the event A is, P(A).

P(A) = 4/52 = 1/13  [There are 4 kings in the 52 cards set]

As the king is withdrawn without replacement there are three kings left only.

And the number of cards left is 51 as well.

Hence,

P(B/A) = 3/51 = 1/17

Hence, the required probability,

P(A∩B) = P(A) × P(B/A) = (1/13) × (1/17)

or, P(A∩B) = 1/221 (ans)

### (ii) Two cards are drawn without replacement from a pack of 52 cards. Find the probability that first is king and the second is an ace.

Solution:

Let the desired events be,

A = first card is king

B = second card is also king

and probability of the event A is, P(A).

P(A) = 4/52 = 1/13  [There are 4 kings in the 52 cards set]

As the king is withdrawn without replacement the number of cards left is 51 and it has 4 aces.

Hence,

P(B/A) = 4/51 [4 aces in 51 cards set]

Hence, the required probability,

P(A∩B) = P(A) × P(B/A) = (1/13) × (4/51)

or, P(A∩B) = 4/663 (ans)

### (iii) Two cards are drawn without replacement from a pack of 52 cards. Find the probability that the first one is a heart and the second one is red.

Solution:

There are 13 heart cards and 26 red cards, if we withdraw a heart card 25 red cards remain.

Let the desired events be,

A = first card is heart.

B = second card is red.

The probability of the event x is P(x).

P(A) = 13/52 = 1/4  [There are 13 hearts in the 52 cards set]

If we withdraw a heart card 25 red cards remain

P(B/A) = 25/51 [26 red cards in 51 cards set]

Hence, the required probability,

P(A∩B) = P(A) X P(B/A) = (1/4) X (25/51)

or, P(A∩B) = 25/204 (ans)

### Question 7: A bag contains 20 tickets, numbered from 1 to 20. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that first will show even number and the second will show anodd number.

Solution:

Let the desired events be,

A = Even number at first draw

B = Odd number at second draw

The probability of the event x is P(x).

P(A) = 10/20 = 1/2 [Any of the 10 even numbers less than 20]

P(B/A) = 10/19 [Any of the 10 remaining odd number tickets out of 19]

The required probability,

P(A∩B) = P(A) × P(B/A)

= (1/2) × (10/19) = 5/19 (ans)

### Question 8: An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement . What is the probability that at least one ball is black?

Solution:

Let the desired events be,

A = Black ball at first draw

B = Black at second draw.

Complement of desired events be,

A’ = A white ball or red ball in first draw.

B’ = A white ball or red ball in second draw.

P(A’) = 7/12 [7 non-black balls out of 12]

P(B’/A’) = 6/11 [6 non-black balls out of 11]

Hence, the required probability,

P(At least one ball is black)

= P(A U B)

= 1 – P(A U B)’

=1 – P(A’ ∩ B’)

=1 – P(A’)P(B/A)’

=1 – (7/12 × 6/11)

=15/22 (ans)

### Question 9: A bag contains 5 white 7 red and 3 black balls. If three balls are drawn one by one without replacement. Find the probability that none of them is red.

Solution:

Let the desired events be,

A = No red ball in first draw.

B = No red ball in second draw.

C = No red ball in the third draw.

P(A) = 8/15 [8 non-red balls of 15]

P(B/A) = 7/14 = 1/2 [7 non-red balls of 14]

P(C/A∩B) = 6/13 [6 non-red balls of 13]

The required probability,

P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)

= 8/15 × 1/2 × 6/13

= 8/65 (ans)

### Question 10: A card is drawn from a well-shuffled deck of 52 cards, then a second card is drawn. Find the probability that thefirst card is heart and thesecond card is diamond. If the first card is not replaced.

Solution:

Let the desired events be,

A = first card is heart

B = second card is diamond

c = 13/52 = 1/4 [13 hearts out of 52 cards]

P(B/A) = 13/51 [13 diamonds out of 51 cards]

The required probability,

P(A∩B)

= P(A) × P(B/A)

=1/4 × 13/51

=13/204 (ans)

### Question 11: An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Solution:

Let the desired events be,

A = first ball is black

B = second ball is black

P(A) = 10/15 = 2/3 [10 black balls of 15]

P(B/A) = 9/14 [9 black balls of 14]

The required probability,

P(A∩B)

= P(A)P(B/A)

= 2/3 × 9/14

= 3/7 (ans)

### Question 12: Three cards are drawn successively, Without replacement from a pack of 52 well-shuffled cards. What is the probability that the first two cards are kings and the third card drawn in an ace?

Solution:

Let the desired events be,

A = first card is king

B = second card is king

C = third card is an ace

P(A) = 4/52 = 1/13 [There are 4 kings and 52 cards remaining]

P(B/A) = 3/51 = 1/17 [There are 3 kings and 51 cards remaining]

P(C/A∩B) = 4/50 = 2/25 [There are 4 aces and 50 cards remaining]

The required probability,

P(A∩B∩C)

= P(A)P(B/A)P(C/A∩B)

= 1/13 × 1/17 × 2/25

= 2/5525 (ans)

### Question 13: A box of oranges is inspected by three randomly selected oranges drawn without replacement. If all three oranges are good, The box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Solution:

Let the desired events be,

A = first orange is good.

B = second orange is good.

C = third orange is good.

P(A) = 12/15 = 4/5 [There are 12 good oranges among 15]

P(B/A) = 11/14 [There are 11 good oranges among 14]

P(C/A∩B) = 10/13 [There are 10 good oranges among 13]

The required probability,

P(A∩B∩C)

= P(A)P(B/A)P(C/A∩B)

= 4/5 × 11/14 × 10/13

= 44/91 (ans)

### Question 14: A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the another without replacement. Find the probability that the balls drawn are white, black and red respectively.

Solution:

Let the desired events be,

A = first ball is white.

B = second ball is black.

C = third ball is red.

P(A) = 4/16 = 1/4 [There are 4 white balls among 16]

P(B/A) = 7/15 [There are 7 black balls among 15]

P(C/A∩B) = 5/14 [There are 5 red balls among 14]

The required probability,

P(A∩B∩C)

= P(A)P(B/A)P(C/A∩B)

= 1/4 × 7/15 × 5/14

= 1/24 (ans)

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