# 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 2 of 4.

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.

Sample= 8+7+6=21.

Probability of selection Red and Green= 8+6=14 / 21 = 2/3.

Not selection of Red and Green. Simple subtracts 1 from 2/3.

Kigele yazidi said:   10 months ago
Total number of balls = 8 + 7 + 6,
=21.

Probability not red or green= number of events/sample space.
So, 7รท21 = 1/3.
(3)

Mangesh Modiraj said:   1 decade ago
@Nandini & @Monica.

In example clearly mentioned that the ball ins neither red nor green.

i.e blue.

We have 7 blue balls.

Therefore n(E) = 7.

Tushar dhok said:   7 years ago
Total no.of ball 8, 7, 6 = 21.

E= event that the ball is neither red nor green.
= event that the ball is red.
n(E) = 8.
P(E) = 8/21.
(1)

Knowledgeispower said:   8 years ago
Using the same problem, solve for the probability of pulling a RED ball on 2 draws without replacing the ball after each draw.

Neither red nor blue.

Then why we are take 7 blue balls only.

Can't understand.

Anitha said:   5 years ago
Total no.of balls =8+7+6=21,
Neither red nor green,
I.e.,we need to take 8,
Therefore p(e)=8/21.

S.Ranga reddy said:   1 decade ago
Hear there are 1 randomly selected means
green 8 blue 6
average of 2colors
so 8+6=14
14/2=7

Boopathy said:   6 years ago
Probability of selection Red or Green= 8+6=14 / 21 = 2/3.

Then how 1/3?
(1)