# Quantitative Aptitude: Probability Questions – Set 23

1. Two dice are rolled randomly. Find the probability to get sum is 10.
2/5
14
2/9
1/12
5/8
Option D
Required = (6 , 4 ) , (4 , 6) , ( 5, 5) = 3
total = 6 * 6 = 36
probability = 3/36 = 1/12

2. Two dices are rolled out together, then what is the probability of getting a number of one dice greater than the number of other dice ?
2/3
1/6
3/8
1/8
5/6
Option E
Non-favorable events = (1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) , (5 , 5) , (6,6) = 6
total = 6 * 6 = 36
probability of non-favorable events = 6/36 = 1/6
probability of favorable events = 1 – 1/6 = 5/6

3. If 2 cards is drawn at randomly from 52 cards, then find the probability of getting both are red cards.
13/28
26/105
28/108
25/102
none of these
Option D
Required = 26C2 = 26 * 25/2 = 13 * 25
total = 52C2 = 52 * 51 / 2 = 26 * 51
probability = 13 * 25 / 26 * 51 = 25 /102

4. In how many ways, we can arrange the letters of the word ‘LIGHT’ ?
84
28
140
160
120
Option E
Ways = 5! = 5 * 4 * 3 * 2 * 1 = 120

5. In how many different ways, we can arrange the letters of the word ‘MOUSE’ , so that the middle position is always occupied by ‘S’ ?
25
24
48
42
68
Option B
Ways = 4! = 4 * 3 * 2 * 1 = 24

6. How many ways 6 books can be selected from 14 different books , if two particular books are always selected ?
495
480
231
384
520
Option A
Available objects = 14 – 2 = 12
Ways = 12C4 = 12 * 11 * 10 * 9/4 * 3 * 2 * 1 = 495

7. A bag contains 2 white balls, 3 pink balls and 2 black balls. 2 balls are drawn randomly. What is the probability that there is no black balls ?
8/21
10/21
1/8
3/5
6/11
Option B
Required = 5C2 = 5*4/2 = 10
total = 7C2 = 7 * 6 / 2 = 21
probability = 10/21

8. When two coins are tossed simultaneously, then find the probability of getting at least one tail.
5
3
3/4
1/4
2/5
Option C
Required = ( 1 head , 1 tail ) , ( 2 tails) = (2C1) + (2C2) = 2 + 1 = 3
total = 2^2 = 4
probability = 3/4

9. A bag contains 4 red balls, 2 blue balls and 2 green balls . If two balls are drawn randomly from the bag , then find the probability of getting both balls of different color.
2/7
5/8
5/7
4
8
Option C
probability = 4C1 * 2C1 / 8C2 + 4C1 * 2C1 / 8C2 + 2C1 * 2C1 / 8C2
(4 * 2 * 2)/ 8 * 7 + (4 * 2 * 2)/ 8 * 7 + (2 * 2 * 2) / 8 * 7 = 2/7 + 2/7 + 1/7 = 5/7

10. A group of students sitting around a rectangular table, find probability of 2 specified students sitting together.
3/8
2/7
8/7
3/8
6/11
Option B
Favorable cases = 6! * 2!
total cases = 7!
probability = 6! * 2! /7! = 2/7

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