## 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**

**Answer with Explanation:**

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**

**Answer with Explanation:**

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**

**Answer with Explanation:**

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**

**Answer with Explanation:**

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 = ^{7}C_{2}/^{12}C_{2} = [(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**

**Answer with Explanation:**

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 **

**Answer with Explanation:**

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**

**Answer with Explanation:**

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**

**Answer with Explanation:**

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

Add the cases

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**

**Answer with Explanation:**

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**

**Answer with Explanation:**

Let 2 events A and B

Odds against A are 1 : 3

So probability of occurrence of A = 3/(1+3) = 3/4. And non-occurrence of A = 1/4

Odds in favor of B are 2 : 5

So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7

Case 1: A occurs and B does not occur

So probability = 3/4 * 5/7 = 15/28

Case 2: B occurs and A does not occur

So probability = 2/7 * 1/4 = 2/28

So probability that one will occur = 15/28 + 2/28 = 17/28