An urn contains 6 red and 4 black balls. They are withdrawn from the urn, one at a time and without replacement. Find the expected number of red balls chosen after the first but before the third black ball has been chosen.

To find the expected number of red balls chosen after the first but before the third black ball has been chosen, we can break down the problem into several cases: Case 1: The first black ball is chosen on the first draw. In this case, there are no red balls chosen between the first and the third black ball because only one black ball has been drawn. Case 2: The first black ball is chosen on the second draw. In this case, the first draw must be a red ball. There are 6 red balls and 9 remaining balls (since one black ball has been removed), so the probability of drawing a red ball on the first draw is 6/9. After the first draw, you are left with 5 red balls and 8 remaining balls (2 black balls and 6 red balls). The probability of drawing a red ball on the second draw is 5/8. So the probability of this case is (6/9) * (5/8). Case 3: The first black ball is chosen on the third draw. In this case, the first two draws must be red balls. The probability of drawing a red ball on the first draw is 6/10, and after the first draw, you are left with 5 red balls and 9 remaining balls (1 black ball and 8 red balls). The probability of drawing a red ball on the second draw is 5/9. After the second draw, you are left with 5 red balls and 8 remaining balls (1 black ball and 7 red balls). The probability of drawing a red ball on the third draw is 5/8. So the probability of this case is (6/10) * (5/9) * (5/8). Now, we can calculate the expected number of red balls chosen between the first and the third black ball by taking the weighted average of these cases: Expected number = (0 * probability in Case 1) + ((1 * probability in Case 2) + (2 * probability in Case 3)) Expected number = (0 * 1) + ((1 * (6/9) * (5/8)) + (2 * (6/10) * (5/9) * (5/8))) Expected number = (0) + ((30/72) + (300/720)) Expected number = (30/72) + (5/12) Now, we can simplify the fraction: Expected number = 5/6 So, the expected number of red balls chosen after the first but before the third black ball has been chosen is 5/6.