Question

a bag contains 5 white and 8 red balls. two drawings of 3 balls are made such that (a) balls are replaced before the second trial and b the balls are not replaced before the second trial . find the probability that the first drawing will give 3 while and the second 3 red with in each case.​

Answer 1

There are total 13 balls out of which 8 are red and 5 are white.

Favourable case of first draw is to get 3 white balls out of 5 white balls.

Probability P1 = 5C3/13C3

If this hapoens then remainings are - 2 white balls and 8 red balls.

Favourable case is to get 3 red balls out of 8 balls.

Probability P2 = 8C3/10C3

Both the events are independent of each other, hence total probability is P1*P2 = (5C3 * 8C3)/(13C3 * 10C3)

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