Question

In a box, there are 9 red, 7 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

• A. 11/21
• B. 1/2
• C. 10/21
• D. 1/3
• E. none

Answer 1

The correct option is D

1/3

Total number of balls = (9 + 7 + 5) = 21.

Let E = event that the ball drawn is neither red nor green.

= event that the ball drawn is blue.

$\therefore$ n(E) = 7.

$\therefore$ P(E)= $\frac{1}{3}$.