Question

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

• A. $\frac{2}{3}$
• B. $\frac{3}{4}$
• C. $\frac{7}{19}$
• D. $\frac{8}{21}$

Answer 1

The correct option is D $\frac{8}{21}$

Total number of balls$=(8+7+6)=21$

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

$=$event that the ball drawn is red.

Therefore, $n\left(E\right)=8$.

$P\left(E\right)=\frac{8}{21}$