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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 & Solution
Correct Answer:
Option
A
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
$$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$
= (8 + 7 + 6)
= 21
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
$$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$
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