# 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.

Chris said:   2 years ago
@All.

Here, You can solve it in two ways.

Method 1:
Total number of balls = 7+8+6= 21.
Neither red nor green, that means it is blue,
Number of blue = 7,
Probability of blue = number of blue ballsÃ· total number of balls.

Probability of blue balls= 7/ 21.

1/3.

Method 2:
Total number of balls = 21.
Number of Red = 8.
Prob of selecting red = 8/21 or.
Number of Green = 6.
Prob of selecting green = 6/21 or 2/7.
Neither Red nor Green = 1-(prob of red+ prob of green).
1 - (8/21 + 6/21).
1 - 14/21 = 7/21 or 1/3.
(10)

There is a die with 10 faces. It is not known that fair or not. 2 captains want to toss die for batting selection. What is the possible solution among the following?
a) If no. is odd it is head, if no. is even it is tail
b) If no. is odd it is tail, if no. is even it is head
c) Toss a die until all the 10 digits appear on top face. And if first no. in the sequence is odd then consider it as tail. If it is even consider it as head.
what is the exact solution

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.

Please all students read question carefully what he ask in question. He said that 8 red, 6 green, 7 blue total=21 probability ?

He said neither from red and green so only one option is 7 blue.

7c1 which is n(a)=7 and n(S)=21c1=21

Formula is prob. p(A)=n(A)/n(S)

7/21 = 1/3.

Total balls are - 21

So the probability of getting one ball is -21c1 = 21.

That one ball is neither red nor green means blue ball.

Posing one blue ball from 7 blue balls is 7c1 = 7.

Our required ans=7/21=1/3.

Hi aruna I think you didn't read the question properly it is clearly mentioned that neither red nor green, so the remaining are blue. Hence we hve to consider blue balls. Hope you got it :).

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?

Shiv said:   3 years ago
Neither red''ï¸ nor green'' means (avoiding red and green) so we left out with blue means the one ball we have to pick up should be blue colour ball so.

P(E) = 7/21 = 1/3.

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.

Angamuthu SURESH said:   1 decade ago
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 red.
Therefore, n(E) = 8.
P(E) = 8/21.