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 blue nor green?

A. $\frac{2}{3}$

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

C. $\frac{3}{7}$

D. $\frac{9}{22}$

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Correct Answer : B.

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) =

\frac{8}{21}

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