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

Solution:

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{aligned} ∴ n (E) = 7 \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{7}{21} = \frac{1}{3} \end{aligned}

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