A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A 10/21

B 11/21

C 1/2

D 2/7

Correct Answer: Option A

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 = \(7C_2\) = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = \(5C_2\)= 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.