# Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 3)
3.
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?
 1 3
 3 4
 7 19
 8 21
 9 21
Explanation:

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) 21 3

Discussion:
36 comments Page 1 of 4.

HOW n(E)=7?

Rajeev R T said:   1 decade ago
Event that the ball is neither red nor green, then you are left out with 7 blue balls, and out 7 blue balls you need one ball.

Then it is 7C1 ways... i.e = 7 ways.

How the n(e) come as 7?

If at least one is blue means what is the answer?

How n(E) =7?

Because neither which is indicates that. So that the possibility for the red and green is not considered. Thus simultaneously the blue ball is taken under consideration.

n(E) =7 because there are 7 blue balls.

Neither red nor blue.

Then why we are take 7 blue balls only.

Can't understand.