# Free Test for Bank PO : Probability

Bank PO exams (IBPS PO and SBI PO) generally ask 5 questions on Probability in the Quantitative Aptitude section. Today we look at 5 questions on Probability and their detailed solutions.

1. A fair dice is rolled 3 times. What is the probability that the sum of numbers in the 3 throws is less than 6?

a. 1/27
b. 5/108
c. 1/36
d. 1/12

2. A number is randomly chosen from 1 to 100(both inclusive). What is the probability that it is divisible by 5 or 7?

a. 3/50
b. 7/25
c. 8/25
d. 17/50

3. Ashwini’s chances of winning against Anand in a chess match is 0.2. In a series of 5 matches, what is the probability that Ashwini will win 3 matches?

a. 0.2304
b. 0.1024
c. 0.004
d. 0.0512

4. The chances of Ajit coming to school on time is 0.8. The probability of Ram coming to school on time is 0.6. The probability of Mysore Shweta coming to school on time in 0.75. The probability that Sneha coming to school on time in 0.4. What is the probability that at least 2 of them come on time on a given day?

a. 0.878
b. 0.732
c. 0.644
d. 0.512

5. There are 3 sacks with 10 blue balls, 5 red balls, and 8 yellow balls respectively. A sack is selected at random, and a ball drawn. What is the probability that the ball selected is blue?

a. 10/23
b. 5/23
c. 10/69
d. 1/3

1. (b)
Minimum sum when 3 dice are rolled = 3 (all of them show 1)
So, we need to consider ways in which sum’s of 3, 4 and 5 show up.
3 – Only 1 way (1,1,1)
4 – (1,1,2), (1,2,1), (2,1,1) – 3 ways
5 – (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1), (3,1,1) – 6 ways
Total outcomes = 6 x 6 x 6 = 216
Favorable outcomes = 1 + 3 + 6 = 10
Probability = 10/216 = 5/108

2. (c)
We have 20 multiples of 5 between 1 to 100
We have 14 multiples of 7 between 1 to 100
We now need to subtract multiples of 35 as they would have been counted twice (once as multiple of 5, and once as multiple of 7)
We have 2 multiples of 35 below 100.
So total numbers divisible by 5 or 7 = 20 + 14 – 2 = 32
Probability = 32/100 = 8/25

3. (d)
Ashwini can win 3 of 5 matches in 5C3 ways.
So, required Probability = 5C3 x (0.2)3 x 0.82
= 10 x 0.008 x 0.64 = 0.0512

4. (a)
Required probability = 1 – no one comes on time – 1 comes on time
no one comes on time = (1-0.8) x (1-0.6) x (1-0.75) x (1-0.4) = 0.012
1 comes on time = 0.8 x ((1-0.6) x (1-0.75) x (1-0.4) ) + 0.6 x ((1-0.8) x (1-0.75) x (1-0.4)) + 0.75 x ((1-0.8) x (1-0.6) x (1-0.4)) + 0.4 x ((1-0.8) x (1-0.6) x (1-0.75))
= 0.048 + 0.018 + 0.036 + 0.008 = 0.11
So, Required probability = 1 – 0.012 – 0.11 = 0.878

5. (d)
For the selected ball to be blue, we need to draw the sack with 10 blue balls.
This has a probability of 1/3.
Note: The number of balls does not matter, as once a sack is selected only a ball of a particular color can be drawn.