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?

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? Correct Answer $$\frac{{8}}{{21}}$$

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(E) = 8
P(E) = $$\frac{{8}}{{21}}$$
Bissoy MCQ

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