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?
Answer: Option A
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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.
So, n(E) = 7.
So, \(P(E) = \frac{n(E)}{n(S)}=\frac{7}{21}=\frac{1}{3}.\)
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