# Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 2)
2.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
 10 21
 11 21
 2 7
 5 7
Explanation:

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 `
 = (7 x 6) (2 x 1)
= 21.

Let E = Event of drawing 2 balls, none of which is blue. n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
 = (5 x 4) (2 x 1)
= 10. P(E) = n(E) = 10 . n(S) 21

Discussion:
106 comments Page 1 of 11.

Ronak Mistry said:   3 weeks ago
Total Balls = 7.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
(1)

Tejeshwari said:   9 months ago
5/7 is it right?

(3)

Lyi said:   2 years ago
The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a non-blue and therefore we have one less ball than 5 to draw from now.

(4)

SRZ said:   2 years ago
P1st (not blue)= 1-P(blue) = 1-2/7 = 5/7.
P2nd(not blue)= 1-P(blue) = 1-2/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
(6)

Thejo said:   2 years ago
We should pick 2 from 7 none of the ball drawn is blue, then the picked balls colours should be red and green. How come it is 10/21?

Here it could be 2C1*3C1 Ã· 7C2 = 6/21 right! Can any1 tell me why it is wrong?
(1)

Kanishk said:   3 years ago
@Biswajit.

Firstly the probability of drawing two blue balls will be 2/7*1/6.
Secondly, it won't work by subtracting from 1 since two blue balls and no blue balls aren't the only two cases. There are other case balls with different colours, one of the colours being blue.

Hope you get it now!
(1)

Biswajit said:   3 years ago
Can anyone tell me if I subtract the probability of getting 2 blue balls i.e 2/7 from 1 then why I can't get the answer?

Sai said:   3 years ago
Total number of balls gives 2+3+2=7 s be the sample space.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.

e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.

Therefore p n(e)= 10/21.

Gufran said:   3 years ago
(2R+3G)c2/7c2.
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).