# 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:
112 comments Page 1 of 12.

Can we do this n different method? finding the possibility that both are blue and finally ans-1.

Its a correct answer and short way.

Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21

Probability that none of ball is blue ---> 1-(1/21) = 20/21

p(a)+p(a(bar)) = 1 .......

This is the the simplest method to find the Probablity...........

Rose were said:   1 decade ago
Why are you multiplying by 6 and by 4?

Rose were said:   1 decade ago

What Kk585 said is not correct.

Becuase, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.

Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.

Therefore total is 11/21.

Hence ans is 1-(11/21) is 10/21.

Can we do it by this way?

3c2+2c2+3c1 2c1
---------------
7c2.

I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?