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. $\frac{3}{4}$
• C. $\frac{7}{19}$
• D. $\frac{8}{21}$
• E. $\frac{9}{21}$

Answer 1

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

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.
$\therefore n\left(E\right)=7$.
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{21}=\frac{1}{3}$.