lakshnachadha wrote:A bag contains 3 white balls, 3 black balls & 2 red balls. One by one three balls are drawn out without replacement. What is the probability that the third ball is red?

3W 3B 2R

In order to get a Red in third pick, the two red balls should not have been picked in first two picks.

[P(Getting Red in first pick) AND P (Getting Something other color in second pick)] OR [P(Getting Something other color in first pick) AND P (Getting Red in second pick)]

= 2/8 * 6/7 + 6/8 * 2/7

= 3/7

Now, P(Getting Red in third pick) = 3/7 * 1/6 = 1/14

[P(Getting Something other color in second pick) AND P (Getting Something other color in second pick)]

= 6/8 * 5/7 = 15/28

Now, P(Getting Red in third pick) = 15/28 * 2/6 = 5/28.

Answer will be 1/14 + 5/28 = 7/28 = 1/4

Is it correct? Let me know.

Probability of drawing red ball on third attempt is same as probability of drawing red ball on first attempt.
Possible outcomes= 2
Total outcomes = 8
Probability = 2/8 = 1/4

beat_gmat_09 wrote:Probability of drawing red ball on third attempt is same as probability of drawing red ball on first attempt.
Possible outcomes= 2
Total outcomes = 8
Probability = 2/8 = 1/4

Phew!! Its that simple... hahaha

Need rest badly. Good Night fellas.

shovan85 wrote:

beat_gmat_09 wrote:Probability of drawing red ball on third attempt is same as probability of drawing red ball on first attempt.
Possible outcomes= 2
Total outcomes = 8
Probability = 2/8 = 1/4

Phew!! Its that simple... hahaha

Need rest badly. Good Night fellas.

C ya 2morrow.. good night..