An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :

Solution

Maths sir · 4 steps

Do the two draws one after the other: after 4 whites leave, the urn has only 11 balls for the second draw.

1. First draw. Choose 4 balls from 15. All 4 are white in C(6, 4) = 15 ways out of C(15, 4) = 1365, so the probability is 15/1365 = 1/91.
2. Urn after first draw. There is no replacement, so 11 balls are left: 2 white and 9 black.
3. Second draw. All 4 are black in C(9, 4) = 126 ways out of C(11, 4) = 330, so the probability is 126/330 = 21/55.

   P = 1/91 × 21/55 = 21/5005 = 3/715

4. So. The required probability is 3/715, which is option (c).

The idea to keep. For draws made one after another without replacement, multiply the probabilities in order, and work each one out from the urn as it stands after the earlier draws.

Answer from the NTA final answer key. Solution by PrepWiser.