91Ó°ÊÓ

In this scenario, we first find the probability of drawing a white ball as the second ball given that a black ball was drawn first, and then given that a white ball was drawn first. We will use the formula for conditional probability: \[P(B_2|A) = \frac{P(B_2 \cap A)}{P(A)}\] where \(B_2\) is the event of drawing a white ball as the second ball, and \(A\) is the event related to the first ball drawn. 1. If a black ball is drawn first, - \(P(\)black ball drawn first\) = \frac{5}{9}\) - \(P(\)white ball drawn as second\ |\ black ball drawn first\) = \frac{4}{8} = \frac{1}{2}\), since there are now 8 balls left in the urn. - \(P(\)white ball drawn as second \cap black ball drawn first\) = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}\) 2. If a white ball is drawn first, - \(P(\)white ball drawn first\) = \frac{4}{9}\) - \(P(\)white ball drawn as second\ |\ white ball drawn first\) = \frac{3}{8}\), since there are now 8 balls left in the urn. - \(P(\)white ball drawn as second \cap white ball drawn first\) = \frac{4}{9} \times \frac{3}{8} = \frac{1}{6}\) We now sum the probabilities from both cases to get the overall probability: \(P(\)white ball as second ball without replacement\) = \frac{5}{18} + \frac{1}{6} = \frac{5 + 3}{18} = \frac{8}{18} = \frac{4}{9}\)

In this scenario, we follow the same steps as in part a, but the probabilities change since the first ball is replaced before the second is drawn. 1. If a black ball is drawn first, - \(P(\)black ball drawn first\) = \frac{5}{9}\) - \(P(\)white ball drawn as second\ |\ black ball drawn first\) = \frac{4}{9}\) (no change in the number of balls since replacement). - \(P(\)white ball drawn as second \cap black ball drawn first\) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}\) 2. If a white ball is drawn first, - \(P(\)white ball drawn first\) = \frac{4}{9}\) - \(P(\)white ball drawn as second\ |\ white ball drawn first\) = \frac{4}{9}\) (no change in the number of balls since replacement). - \(P(\)white ball drawn as second \cap white ball drawn first\) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}\) We now sum the probabilities from both cases to get the overall probability: \(P(\)white ball as second ball with replacement\) = \frac{20}{81} + \frac{16}{81} = \frac{36}{81} = \frac{4}{9}\)

The probability of drawing a white ball as the second ball is: a. Without replacement: \(\frac{4}{9}\) b. With replacement: \(\frac{4}{9}\)