In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A. $$\frac{{1}}{{3}}$$

B. $$\frac{{3}}{{4}}$$

C. $$\frac{{7}}{{19}}$$

D. $$\frac{{8}}{{21}}$$

E. $$\frac{{9}}{{21}}$$

Answer: Option A

Solution(By Examveda Team)

Total number of balls
= (8 + 7 + 6)
= 21
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
$$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$