A bag holds 5 red and 3 blue balls. Two balls are drawn. Let p be the probability that both are red when the first ball is put back before the second draw, and let q be the probability that both are red when it is not put back. Find p − q.

With replacement each draw sees 5 red out of 8, so p = (5/8)(5/8) = 25/64. Without replacement the second draw sees 4 red out of the 7 that remain, so q = (5/8)(4/7) = 5/14. Writing both over 448 gives p = 175/448 and q = 160/448, so p − q = 15/448. The difference is not zero: removing a red ball makes a second red slightly less likely, so p is the larger of the two.