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

$\frac{{}^{6}C_4}{{}^{15}C_4}=\frac{1}{91}$

The balls are not replaced.

Now, the probability that second draw give four black balls

$\frac{{}^{9}C_4}{{}^{11}C_4}=\frac{21}{55}$

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

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

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