An urn contains 4 white balls and 6 black balls. A ball is chosen at random and its color noted. The ball is then replaced, along with 3 more balls of the same color (so that there are now 13 balls in the urn). Then another ball is drawn at random from the urn. (a) Find the chance that the second ball drawn is white.
(b) Given that the second ball drawn is white, what is the probability that the first ball drawn is black?
(c) Suppose the original contents of the urn are w white and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. In part (a), w was 4, b was 6, and d was 3. Show that the chance that the second ball drawn is white is w/w+b
b. Note that the answer does not depend on the value of d. Why is that?