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 :

Total balls: 6 white + 9 black = 15.

First draw: 4 white balls from 6 white: $$\binom{6}{4} / \binom{15}{4}$$.

After first draw: 2 white + 9 black = 11 balls remain.

Second draw: 4 black balls from 9: $$\binom{9}{4} / \binom{11}{4}$$.

$$P = \frac{\binom{6}{4}}{\binom{15}{4}} \times \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{15}{1365} \times \frac{126}{330} = \frac{15 \times 126}{1365 \times 330}$$

$$= \frac{1890}{450450} = \frac{3}{715}$$

The answer corresponds to Option (3).