# Quantitative Aptitude: Probability Questions – Set 3

Probability Questions for Bank PO Exams – SBI PO, NIACL, NICL, BoB PO, IBPS PO, and other exams

1. There are 100 tickets in a box numbered 1 to 100. 3 tickets are drawn at one by one. Find the probability that the sum of number on the tickets is odd.
A) 2/7
B) 1/2
C) 1/3
D) 2/5
E) 3/7
Option B
Solution:
There will be 4 cases
Case 1: even, even, odd
Prob. = 1/2 × 1/2 × 1/2
Case 2: even, odd, even
Prob. = 1/2 × 1/2 × 1/2
Case 3: odd, even, even
Prob. = 1/2 × 1/2 × 1/2
Case 4: odd, odd, odd
Prob. = 1/2 × 1/2 × 1/2
Add all the cases, required prob. = 1/2
2. There are 4 green and 5 red balls in first bag. And 3 green and 5 red balls in second bag. One ball is drawn from each bag. What is the probability that one ball will be green and other red?
A) 85/216
B) 34/75
C) 95/216
D) 35/72
E) 13/36
Option D
Solution:
Case 1:first green, second red
Prob. = 4/9 × 5/8 = 20/72
Case 2:first red, second green
Prob. = 5/9 × 3/8 = 15/72
3. A bag contains 2 red, 4 blue, 2 white and 4 black balls. 4 balls are drawn at random, find the probability that at least one ball is black.
A) 85/99
B) 81/93
C) 83/99
D) 82/93
E) 84/99
Option A
Solution:
Prob. (At least 1 black) = 1 – Prob. (None black)
So Prob. (At least 1 black) = 1 – (8C4/12C4) = 1 – 14/99
4. Four persons are chosen at random from a group of 3 men, 3 women and 4 children. What is the probability that exactly 2 of them will be men?
A) 1/9
B) 3/10
C) 4/15
D) 1/10
E) 5/12
Option B
Solution:
2 men means other 2 woman and children
So prob. = 3C2 × 7C2 /10C4 = 3/10
5. Tickets numbered 1 to 120 are in a bag. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 8/15
B) 5/16
C) 7/15
D) 3/10
E) 13/21
Option C
Solution:
Multiples of 3 up to 120 = 120/3 = 40
Multiples of 5 up to 120 = 120/5 = 24 (take only whole number before the decimal part)
Multiple of 15 (3×5) up to 120 = 120/15 = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = 56/120 = 7/15
6. There are 2 people who are going to take part in race. The probability that the first one will win is 2/7 and that of other winning is 3/5. What is the probability that one of them will win?
A) 14/35
B) 21/35
C) 17/35
D) 19/35
E) 16/35
Option D
Solution:
Prob. of 1st winning = 2/7, so not winning = 1 – 2/7 = 5/7
Prob. of 2nd winning = 3/5, so not winning = 1 – 3/5 = 2/5
So required prob. = 2/7 * 2/5 + 3/5 * 5/7 = 19/35
7. Two cards are drawn at random from a pack of 52 cards. What is the probability that both the cards drawn are face card (Jack, Queen and King)?
A) 11/221
B) 14/121
C) 18/221
D) 15/121
E) 14/221
Option A
Solution:
There are 52 cards, out of which there are 12 face cards.
So probability of 2 face cards = 12C2/52C2 = 11/221
8. A committee of 5 people is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have less number of boys than girls?
A) 7/12
B) 7/15
C) 6/13
D) 5/14
E) 7/13
Option D
Solution:
Case 1: 1 boy and 4 girls
Prob. = 5C1 × 4C4/9C5 = 5/146
Case 2: 2 boys and 3 girls
Prob. = 5C2 × 4C3/9C5 = 40/126
Add the two cases = 45/126 = 5/14
9. A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at random. What is the probability that they are not of same color?
A) 5/11
B) 14/33
C) 2/5
D) 6/11
E) 2/3
Option E
Solution:
Three cases
Case 1: one red, 1 blue
Prob = 2C1 × 4C1 / 12C2 = 4/33
Case 2: one red, 1 white
Prob = 2C1 × 6C1 / 12C2 = 2/11
Case 3: one white, 1 blue
Prob = 6C1 × 4C1 / 12C2 = 4/11
10. A bag contains 5 blue balls, 4 black balls and 3 red balls. Six balls are drawn at random. What is the probability that there are equal numbers of balls of each color?
A) 11/77
B) 21/77
C) 22/79
D) 13/57
E) 15/77
Option E
Solution:
5C2× 4C2× 3C2/ 12C6

## 13 Thoughts to “Quantitative Aptitude: Probability Questions – Set 3”

1. •?((¯°·._.• ąkduuu •._.·°¯))؟•

probability :||
i skip

2. Preeti

2nd q? mine ans is 35/72

3. Preeti

4/9*5/8 + 5/9*3/8 = 35/72

1. yes yes

5 ki jagah 4 se calculate kar diya

1. Vision

So Prob. (At least 1 black) = 1 – (8C4/12C4) = 14/99
So Prob. (At least 1 black) = 1 – (8C4/12C4) = 1-14/99=85/99 hoga

1. Vision

qs no 3

4. jaga

thank u mam…:)

5. vamsy

Thank you mam

6. Ayushi

8th ka solution correct h ky

1. prob. that sum of number on the tickets is odd to be found

7. Pawan Charan

by the way thank you very much for the questions they are really good and helpful tooo

8. kumkum ahuja

ty az:)