an urn contains 6 white and 9 blacks balls. if 4 balls are to be randomly selected without replacement, what is the probabilty that the first 2 selected are white and the last 2 black?

Question

anonymous · Answer

The probability that the first one is white is 2/5.  The probability that the second is white, given that the first was, is 2/5*5/14.  The chance the third ball is black, given the first two were white, is 2/5*5/14*9/13.  The chance the first two are white, then the next two are black is 2/5*5/14*9/13*2/3

anonymous · Answer

About 18/273, which is a little less than 7%.

anonymous · Answer

here ,first find the probability to get first two selected ball is white.
P(w)=(6C2/15C2).
And then find the probability to get second two ball is black.
P(b)=(9C2/13C2)..
Here follow multiplication rule.
Required probability=P(w)*P(b)...find it

anonymous · Answer

Required probability=18/273...

anonymous · Answer

Nice approach.  A little more sophisticated than my basic counting, but same answer.