# Probability | Answer with Explanations

## Probability Problem with Explanations

For all competitive exams Probability is an important chapter. From Probability chapter at least 1-2 problems are appear in all competitive exams like SSC and bank exams (SBI and IBPS-clerk and PO). This section will teach you to understand Probability problems with explanation and short cut to solve problems quickly.

### Probability – Study material for all competitive exams

Probability is an important topic for all competitive exams. This Probability section will teach you how to solve Probability questions. Some important and repeated questions are covered this chapter.

### Problems – Answers with Explanation:

1) From a pack of 52 cards, 1 card is chosen at random. What is the probability of the card being diamond or queen?
A) 2/7
B) 6/15
C) 4/13
D) 1/8
E) 17/52

In 52 cards, there are 13 diamond cards and 4 queens.

1 card is chosen at random
For 1 diamond card, probability = 13/52
For 1 queen, probability = 4/52
For cards which are both diamond and queen, probability = 1/52
So required probability = 13/52 + 4/52 – 1/52 = 16/52 = 4/13

2) From a pack of 52 cards, 1 card is drawn at random. What is the probability of the card being red or ace?
A) 5/18
B) 7/13
C) 15/26
D) 9/13
E) 17/26

In 52 cards, there are 26 red cards and 4 ace and there 2 such cards which are both red and ace.

1 card is chosen at random
For 1 red card, probability = 26/52
For 1 ace, probability = 4/52
For cards which are both red and ace, probability = 2/52
So required probability = 26/52 + 4/52 – 2/52 = 28/52 = 7/13

3) There are 250 tickets in an urn numbered 1 to 250. One ticket is chosen at random. What is the probability of it being a number containing a multiple of 3 or 8?
A) 52/125
B) 53/250
C) 67/125
D) 101/250
E) 13/25

Multiples of 3 up to 250 = 250/3 = 83 (take only whole number before the decimal part)
Multiples of 8 up to 250 = 250/3 = 31
Multiples of 24 (3×8) up to 250 = 250/24 = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = 104/250 = 52/125

4) There are 4 white balls, 5 blue balls and 3 green balls in a box. 2 balls are chosen at random. What is the probability of both balls being non-blue?
A) 23/66
B) 5/18
C) 8/21
D) 7/22
E) 1/3

Both balls being non-blue means both balls are either white or green

There are total 12 balls (4+3+5)
and total 7 white + green balls.
So required probability = 7C2/12C2 = [(7*6/2*1) / (12*11/2*1)] = 21/66 = 7/22

5) There are 4 white balls, 3 blue balls and 5 green balls in a box. 2 balls are chosen at random. What is the probability that first ball is green and second ball is white or green in color?
A) 1/3
B) 5/18
C) 1/2
D) 4/21
E) 11/18

There are total 4+3+5 = 12 balls
Probability of first ball being green is = 5/12
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or green is 8/12 = 2/3
So required probability = 5/12 * 2/3 = 5/18

6) There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?
A) 52/125
B) 53/150
C) 17/50
D) 37/150
E) 32/75

Multiples of 3 up to 150 = 150/3 = 50
Multiples of 7 up to 150 = 150/7 = 21 (take only whole number before the decimal part)
Multiples of 21 (3×7) up to 150 = 150/21 = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = 64/150 = 32/75

7) There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a ticket which has a prime number on it?
A) 3/55
B) 5/58
C) 8/21
D) 16/55
E) 4/13

Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 41, 47, 53.
So probability = 16/55

8) A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue?
A) 61/110
B) 59/108
C) 45/134
D) 53/108
E) 57/110

Case 1: Ball from first bag is white, from another is blue
So probability = 4/9 * 7/12 = 28/108
Case 1: Ball from first bag is blue, from another is white
So probability = 5/9 * 5/12 = 25/108
So required probability = 28/108 + 25/108 = 53/108

9) The odds against an event are 2 : 3 and the odds in favor of another independent event are 3 : 4. Find the probability that at least one of the two events will occur.
A) 11/35
B) 27/35
C) 13/35
D) 22/35
E) 18/35

Let 2 events A and B
Odds against A are 2 : 3
So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5
Odds in favor of B are 3 : 4
So probability of occurrence of B = 3/(3+4) = 3/7. And non-occurrence of B = 4/7
Probability that at least one occurs
Case 1: A occurs and B does not occur
So probability = 3/5 * 4/7 = 12/35
Case 2: B occurs and A does not occur
So probability = 3/7 * 2/5 = 6/35
Case 3: Both A and B occur
So probability = 3/5 * 3/7 = 9/35
So probability that at least 1 will occur = 12/35 + 6/35 + 9/35 = 27/35

10) The odds against an event are 1 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.
A) 17/28
B) 5/14
C) 11/25
D) 9/14
E) 19/28