An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and second draw gives 3 black balls.
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let P(A) be selection of 3 white balls and P(B) be selection of 3 black balls. 

Lets us assume that in the first draw all the 3 balls are white.

So the probability for getting 3 black balls in the Second draw given 3 white balls obtained in first draw is P(B/A)=P(A/B)*P(B)/P(A)

But we  have assumed the first draw. so the probablity of getting 3 white balls is P(A)

So total probability is P(B/A)*P(A)=(P(A/B)*P(B)/P(A))*P(A)=P(A/B)*P(B)=(5c3/10c3)*P(8c3/13c3)=7/429