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
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
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
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
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
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
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
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
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
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
