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?
Solution
Correct Answer: Option D
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn in neither blue nor green.
= even that the ball drawn in blue.
∴ n(E) = 8
∴ P(E) = n(E)/n(S)
= 8/21
Let E = event that the ball drawn in neither blue nor green.
= even that the ball drawn in blue.
∴ n(E) = 8
∴ P(E) = n(E)/n(S)
= 8/21
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