91Ó°ÊÓ

When drawing balls with replacement, the total number of possibilities remains the same for each draw, and the events are independent. First, we calculate the denominator: the probability that the sample drawn contains exactly 3 white balls. There are 4 balls to be drawn, so the number of ways to draw exactly 3 white balls and 1 black ball is \(\binom{4}{3} = 4\). Each draw has a probability of \(\frac{8}{12}\) for a white ball and \(\frac{4}{12}\) for a black ball. Thus, the probability of drawing exactly 3 white balls is: \[P(3W) = \binom{4}{3}\left(\frac{8}{12}\right)^3\left(\frac{4}{12}\right) \] Next, we calculate the numerator: the probability that the first and third balls drawn will be white, given that there are exactly 3 white balls in total. Since the first and third balls are white, there are only two possibilities for the remaining 2 positions: (WWBW) and (WBBW). The probabilities for these cases are: \[(WWBW): \left(\frac{8}{12}\right)^3\left(\frac{4}{12}\right)\] \[(WBBW): \left(\frac{8}{12}\right)^2\left(\frac{4}{12}\right)^2 \] Now, we sum these probabilities: \[P(1^{st}W \cap 3^{rd}W | 3W) = \left(\frac{8}{12}\right)^3\left(\frac{4}{12}\right) + \left(\frac{8}{12}\right)^2\left(\frac{4}{12}\right)^2\] We can now calculate the conditional probability: \[P(1^{st}W \cap 3^{rd}W | 3W) = \frac{ \left(\frac{8}{12}\right)^3\left(\frac{4}{12}\right) + \left(\frac{8}{12}\right)^2\left(\frac{4}{12}\right)^2}{ \binom{4}{3}\left(\frac{8}{12}\right)^3\left(\frac{4}{12}\right)}\] By simplifying, we get: \[P(1^{st}W \cap 3^{rd}W | 3W) = \frac{ \left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right) + \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2}{4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)}\] After simplifying further, we find: \[P(1^{st}W \cap 3^{rd}W | 3W) = \frac{1}{2}.\]

The probability of drawing exactly 3 white balls without replacement is given by: \[ P(3W) = \frac{\binom{8}{3}\binom{4}{1}}{\binom{12}{4}}\] There are two possibilities for the sequence of balls such that the first and third balls drawn are white, with 3 white balls in total: (WWBW): \[\frac{\binom{7}{1}\binom{4}{1}}{\binom{11}{3}}\] (WBBW): \[\frac{\binom{7}{0}\binom{4}{2}}{\binom{11}{3}}\] Adding the probabilities for both sequences: \[P(1^{st}W \cap 3^{rd}W | 3W) = \frac{\frac{\binom{7}{1}\binom{4}{1}}{\binom{11}{3}} + \frac{\binom{7}{0}\binom{4}{2}}{\binom{11}{3}}}{\frac{\binom{8}{3}\binom{4}{1}}{\binom{12}{4}}}\] Simplifying the expression, we obtain: \[P(1^{st}W \cap 3^{rd}W | 3W) = \frac{13}{22}\] The conditional probabilities in each case are: 1. With replacement: \(\frac{1}{2}\) 2. Without replacement: \(\frac{13}{22}\)