Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating question sample in **Probability**, which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!

**A box contains tickets numbered from 1 to 24. 3 tickets are to be chosen to give 3 prizes. What is the probability that at least 2 tickets contain a number which is multiple of 3?**

A) 35/256

B) 33/220

C) 63/253

D) 43/190

E) 59/253

**C) 63/253**

Explanation:

From 1 to 24, there are 8 numbers which are multiple of 3

Case 1: 2 are multiple of 3, and one any other number from (24-8) = 16 tickets

^{8}C_{2}*^{16}C_{1}/^{24}C_{3}= 56/253

Case 2: all are multiples of 3.

^{8}C_{3}/^{24}C_{3}= 7/253

Add both cases.

**A box contains 6 blue, 5 green and 4 red balls. Two balls are drawn at random. What is the probability that there is no red ball?**

A) 3/30

B) 11/21

C) 5/18

D) 11/23

E) None of these

**B) 11/21**

Explanation:

Total balls = 15

Not red means green or blue i.e. any of (5+6) = 11 balls

So prob. =^{11}C_{2}/^{15}C_{2}

**From a pack of 52 cards, 2 cards are drawn at random. What is the probability that both cards are black card or heart card?**

A) 31/102

B) 21/73

C) 1/5

D) 17/100

E) 3/10

**A) 31/102**

Explanation:

Prob. of black card:

^{26}C_{2}/^{52}C_{2}= 25/102

Prob. of heart card:

^{13}C_{2}/^{52}C_{2}= 3/51

Add both cases.

**Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are jacks?**

A) 62/221

B) 21/312

C) 5/21

D) 55/221

E) None of these

**D) 55/221**

Explanation:

There are total 26 cards black, and 4 jacks in which 2 are black jacks

So case 1: both are black

^{26}C_{2}/^{52}C_{2}

case 2: both are jack

^{4}C_{2}/^{52}C_{2}

Add both cases.

But now 2 black jacks have been added in both cases, so subtracting their prob. :^{2}C_{2}/^{52}C_{2}

So 325/1326 + 6/1326 – 1/1326

**From a group of 3 men, 4 women and 2 children, 4 people are to be chosen to form a committee. What is the probability that the committee contains 1 each of men, women and children?**

A) 4/15

B) 12/21

C) 4/19

D) 11/31

E) None of these

**B) 12/21**

Explanation:

Case 1: Prob. when 2 men, 1 woman and 1 child

^{3}C_{2}*^{4}C_{1}*^{2}C_{1}/^{9}C_{4}= 4/21

Case 2: Prob. when 1 man, 2 women and 1 child

^{3}C_{1}*^{4}C_{2}*^{2}C_{1}/^{9}C_{4}= 2/7

Case 3: Prob. when 1 man, 1 woman and 2 children

^{3}C_{1}*^{4}C_{1}*^{2}C_{2}/^{9}C_{4}= 2/21

**A box contains 25 bulbs out of which 5 are defective. 3 bulbs are to be delivered to a customer. What is the probability that he get one defective bulb?**

A) 19/46

B) 25/51

C) 44/77

D) 21/46

E) None of these

**A) 19/46**

Explanation:

^{5}C_{1}*^{20}C_{2/25C3 }**There are 4 red balls, 5 white and 3 green balls in a basket. 3 balls are chosen at random. What is the probability that there is at most 1 green ball?**

A) 13/40

B) 48/55

C) 25/68

D) 8/33

E) 9/19

**B) 48/55**

Explanation:

Case 1: 0 green ball means all three red or white balls

^{9}C_{3}/^{12}C_{3}

Case 2: 1 green ball and two red or white balls

^{9}C_{2}*^{3}C_{1}/^{12}C_{3}

Add both cases.

**A bag contains 3 red, 4 green and 3 yellow balls. If 2 balls are drawn at random, what is the probability that they are of different color?**

A) 9/16

B) 4/15

C) 11/15

D) 5/11

E) None of these

**C) 11/15**

Explanation:

This will be = 1- prob.(both are same in color)

Prob. of both same in color = [^{3}C_{2}+^{4}C_{2}+^{3}C_{2}]/^{10}C_{2}= 12/45

So required prob. = 1 – 12/45

**There are 4 black balls and 6 white balls. 2 balls are drawn one by one without replacement. What is the probability that the balls are same in color?**

A) 9/21

B) 8/17

C) 5/14

D) 7/15

E) 9/19

**D) 7/15**

Explanation:

When both black, prob. = 4/10 * 3/9

When both white, prob. = 6/10 * 5/9

Add both cases.

**A bag contains 5 red balls and 4 green balls. What is the probability that both balls are same in color?**

A) 5/11

B) 4/9

C) 3/13

D) 6/17

E) None of these

**B) 4/9**

Explanation:

Case 1: both red

^{5}C_{2}/^{9}C_{2}

Case 2: both green

^{4}C_{2}/^{9}C_{2}

Add both cases

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