An urn contains 6 white and 9 black balls. Two successive draws of...

Total number of balls in the bag $= 6 + 9 = 15$.

Number of white balls $= 6$.

Number of black balls $= 9$.

Four balls are drawn at random at a time.

The probability that first draw give four white balls

$\dfrac{C4 6}{C4 15} = \dfrac{1}{91}$

The balls are not replaced.

Now, the probability that second draw give four black balls

$\dfrac{C4 9}{C4 11} = \dfrac{21}{55}$

The probability that the first draw give four white balls  and second draw give four black balls

$= \dfrac{1}{91} \times \dfrac{21}{55} = \dfrac{3}{715}$

Hence, the required probability is $\dfrac{3}{715}$.