Question

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. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{3}$

Total number of balls $=(8+7+6)=$ $n\left(S\right)=21$

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

$n\left(E\right)=7.$

Hence the probability that it is neither red nor green, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{21}=\frac{1}{3}$