ok so the answer to this problem:
A bag contains five white balls and four black balls. Your goal is to draw two black balls.
You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)
case 1: getting 2 black balls on 1st try = 4/9*3/8= 1/6
case 2: getting 1 black and 1 white: the prob that you chose one black ball and one white ball is \(\dfrac{4 \cdot 5}{\binom{9}{2}} = \dfrac{5}{9}\). but then, you'd have to put back the white ball, so the bag currently contains three black balls and five white balls. so the prob that you draw a black from there is \(\frac{3}{8}\).
case 3: getting 2 white balls: the probability that you chose two white balls is \(\dfrac{\binom{5}{2}}{\binom{9}{2}}= \dfrac{5}{18}\). like in the last case, you'd have to put back both white balls, so the bag contains four black balls and five white balls. The probability that you draw two black balls is then \(\frac{1}{6}\).
so, the answer is \(\frac{1}{6} + \frac{5}{9} \cdot \frac{3}{8} + \frac{5}{18} \cdot \frac{1}{6} = \boxed{\frac{91}{216}}.\)
.
