# Quantitative Aptitude: Probability Questions – Set 20

1. A bag contains 8 green balls and 12 blue balls. If a ball is drawn randomly, then find the probability of getting either green ball or blue ball.
2
3
2/5
2/3
1
Option E
required = (8C1) + (12C1) = 20
total = (20C1) = 20
probability = 20/20 = 1

2. If two cards are drawn randomly from a pack of 52 cards. Then what is the probability of both cards are red cards ?
28/91
23/108
25/102
8
17
Option C
required = (26C2) = 13 * 25
Total = (52C2) = 26 * 51
probability = (13 * 25)/(26 * 51) = 25/102

3. How ,any 4-digit numbers can be formed from the digits 6, 2, 5, 7, 3, which are divisible by 5 and none of the digits is repeated ?
140
220
28
24
96
Option D
A number to be divisible by 5, unit digit must be 0 or 5.
So, in 4-digit numbers unit place is required to be filled by digit 5 and rest three places of required 4-digit numbers will be filled by rest 4 digit.
numbers of ways = 4! = 4 * 3 * 2 * 1 = 24

4. Out of 52 playing cards two cards is picked randomly, then what is the probability of getting that one red king and one black queen ?
2/663
5/13
8/58
5
14
Option A
required = (2C1) * (2C1) = 4
Total = (52C2) = 26 * 51
probability = 4/(26*51) = 2/663

5. A bag contains 2 red pens, 6 blue pens and 4 green pens. Find the probability of getting two same color pens from the bag ?
5/8
2/4
1/3
5/6
4
Option C
required = (2C2) + (6C2) + (4C2)
= 1 + 15 + 6 = 22
total = (12C2) = 66
probability = 22/66 = 1/3

6. A basket contains 3 green, 4 orange and 5 pink balls. If two balls are drawn randomly from the basket, then what the probability of selecting that both balls are of different colors ?
4/65
57/68
8/45
47/66
4/15
Option D
possibility of both balls are different colors = ( 1 green and 1 orange) , (1 orange and 1 pink ), (1 green and 1 pink)
required = (3C1) * (4C1) + (4C1) * (5C1) + (3C1) * (5C1)
= 12 + 20 + 15 = 47
total = (12C2) = 66
probability = 47/66

7. Sunil threw two dices together. What is the probability of getting that the sum of the two outcomes is 4 ?
1/8
1/12
4
5
2
Option B
Possibility to get = (2, 2), (3, 1), (1, 3)
required = 3
total = 36
probability = 3/36 = 1/12

8. When three coins are tossed simultaneously, then find the probability of getting at least one tale.
7/8
5/8
1/4
4/15
2
Option A
possibility = 1 tale or 2 tale or 3 tale
required = (3C1) + (3C2) + (3C3)
= 3 + 3 + 1 = 7
total = 2^3 = 8
probability = 7/8

9. Four boys and 2 girl sit in a row for presenting their project paper, then what is the probability that they will sit in alternate position ?
1/6
2/3
1/8
1/15
2/5
Option D
probability = 4! * 2!/6! = 1/15

10. Probability of a question solved by A and B is 1/5 and 1/4 respectively, then find the probability that at least one of them will solve the questions ?
5/6
1/4
11/25
2/5
5/9
Option D
possibility = 1 – probability of question not solved by anyone
probability = 1 – (1 – 1/5) * (1 – 1/4)
= 1-(4/5) * (3/4)
= 2/5