# An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first

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An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.

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Given: An urn containing 5 white and 8 black balls .Each trial is independent of the other trial

To find: the probability that the first draws gives 3 white and the second draw gives 3 black balls.

Let , success in the first draw be getting 3 white balls.

Now , the Probability of success in the first trial is

P1(success) = $\frac{5{c_3}}{13c_3}=\frac{10}{286}=\frac{5}{143}$

Let success in the second draw be getting 3 black balls.

Probability of success in the second trial without replacement of the first draw is given by

P2(success) = $\frac{8{c_3}}{10c_3}=\frac{56}{120}=\frac{7}{15}$

Hence , the probability that the first draws gives 3 white and the second draw gives 3 black balls,with each trial being independent is given by 