# 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.

Tharunteja said:   6 years ago
Guys it may be a wrong answer.

Total no.of balls=(8+7+6)=21.
let E = event that the ball drawn is neither red nor green.
=event that the ball drawn is red/

Therefore n(E) = 8 (given neither red nor green) make a note this point.
therefore p(E) = 8/21.

SRINU said:   10 years ago
Total balls-21.
7C1 = 7.
7/21.

Jaiprakash said:   8 years ago
@Obiano.

Ramachandran Nichite said:   8 years ago
Hi @Obiano.

Total ball = 5 + 3 + 2 = 10.

Choose yellow ball = 5/10.

Obiano kingseley said:   8 years ago
A bag contains 5yellow balls, 3green balls, and 2red balls, a ball is chosen at random from the bag. What is the probability that the chosen ball is yellow?

Faffy said:   8 years ago
Does this mean that we are excluding green and red?

Knowledgeispower said:   9 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.

Otuoboizy said:   9 years ago
Total numbers of balls is 21 and cl is 7.

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

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.