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 1

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.

 n(E) = 7.

 P(E) =n(E)=7=1.n(S)213

hope this will helps u

Answer 2

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.
Therefore, n(E) = 7
P(E) = 7/21.