RK73 wrote:

Stop! i stuck!

let's do it slowly -

when we choose three balls, we have 6c3 possibilities.

from these choices we have 3c1 outcomes with blue one.

where I am wrong?

There are 6c3 or 20 possible combinations of 3 balls.

To calculate the number of possible arrangement of 3 balls which include a blue ball assume that the arrangement has a blue ball. Now calculate the number of ways that 2 balls can be combined of the 5 left to accompany the blue one. This is 5c2 = 10. Hence, the probability is 1/2.

Another way to do this is say:

Suppose I pull the balls out 1 by 1 (w/o replacement). For the 3 balls to have the blue one, it must be the 1st, 2nd, or 3rd ball.

Pr(1st) = 1/6

pr(2nd) = 5/6*1/5 = 1/6

pr(3rd) = 5/6*4/5*1/4 = 1/6

These are mutually exclusive events, so we can add them together to get a total probability of 1/2.

_________________

Best,

AkamaiBrah

Former Senior Instructor, Manhattan GMAT and VeritasPrep

Vice President, Midtown NYC Investment Bank, Structured Finance IT

MFE, Haas School of Business, UC Berkeley, Class of 2005

MBA, Anderson School of Management, UCLA, Class of 1993