I agree with your rationale for part a).
As for part b), you can apply the same logic of breaking each draw down. For the first two draws, you would have the same probability of \(\frac{2}{9}\) for drawing two black balls. The probability of having one or more white balls at this point is a complement of your desired event of having two black balls, so the probability of having one or more white balls is \(1 - \frac{2}{9} = \frac{7}{9}\). In other words, you will be redrawing \(\frac{7}{9}\) of the time. The probability of drawing two white balls would be the same as drawing two black balls because there are 5 of each, thus P(both white balls) = \(\frac{2}{9}\). Thus, the only other possibility of drawing exactly one white ball and one black ball would be \(1 - \frac{2}{9} - \frac{2}{9} = \frac{5}{9}\).
Say you drew one white ball in your first drawing. This happens \(\frac{5}{9}\) of the time. If you put the white ball back, your probability of drawing a second black ball from the 9 with 4 black balls would be \(\frac{4}{9}\), so this probability is \(\frac{5}{9}*\frac{4}{9} = \frac{20}{81}\).
Say you drew two white balls in your first drawing. This happens \(\frac{2}{9}\) of the time. If you put both balls back, your probability of drawing two black balls would be the same as the initial drawing, since you have 5 black and 5 white in the bag now. Thus, this probability is \(\frac{2}{9}*\frac{2}{9}=\frac{4}{81}\).
Using the sum rule, as these are all disjoint events, and including the probability we chose two black balls on the first draw, we arrive at the probability being \(\frac{2}{9} + \frac{20}{81} + \frac{4}{81} = \frac{14}{27}\). This could be wrong I gotta get to class lmao
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