An urn contains 6 white and 9 black balls.

Question

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 :

Accepted Answer

$$P(\text{first 4 white}) = \frac{{}^6C_4}{{}^{15}C_4} = \frac{15}{1365}$$

Given first draw all white, 11 balls remain (9 black, 2 white).

$$P(\text{second 4 black}\mid\text{first 4 white}) = \frac{{}^9C_4}{{}^{11}C_4}=\frac{126}{330}$$

$$P = \frac{15}{1365}\times\frac{126}{330} = \frac{3}{715}$$

Solutions · 1