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?

Answer: Option A

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
$$\begin{array}{rl}& \therefore n(E)=7\\ & \therefore P(E)=\frac{n(E)}{n(S)}=\frac{7}{21}=\frac{1}{3}\end{array}$$