An urn contains 5 white and 8 white black balls.Two successive drawing of three balls at a time aremade such that the balls are not replaced before thesecond draw. Find the probability that the first drawgives 3 white balls and second draw gives 3 blackballs.

The problem seems to have a typo. It should probably read "An urn contains 5 white and 8 black balls." Let's solve it with this assumption.

Step 1: Identify the total number of balls. We have 5 white balls and 8 black balls, so the total is 13 balls.

Step 2: Calculate the probability of drawing 3 white balls in the first draw. This is a combination problem, so we use the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.

The probability is C(5, 3) / C(13, 3). Calculate C(5, 3) = 5! / [3!(5-3)!] = 10. Calculate C(13, 3) = 13! / [3!(13-3)!] = 286. So, the probability of drawing 3 white balls in the first draw is 10/286 = 0.034965.

Step 3: After the first draw, we have 2 white balls and 8 black balls left, so the total is 10 balls.

Step 4: Calculate the probability of drawing 3 black balls in the second draw. This is also a combination problem.

The probability is C(8, 3) / C(10, 3). Calculate C(8, 3) = 8! / [3!(8-3)!] = 56. Calculate C(10, 3) = 10! / [3!(10-3)!] = 120. So, the probability of drawing 3 black balls in the second draw is 56/120 = 0.466667.

Step 5: Since the two events are independent, the probability of both events happening is the product of their individual probabilities. So, the probability that the first draw gives 3 white balls and the second draw gives 3 black balls is 0.034965 * 0.466667 = 0.016316.