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.
