Let the event that the
$\rightarrow1^{st}$ is drawn is white be $W_{1}$
$\rightarrow2^{nd}$ ball is white be $W_{2}$
$\rightarrow3^{rd}$ ball is black be $B_{3}$
$\rightarrow4^{th}$ ball is black be $B_{4}$.
$P(W_{1}\cap W_{2}\cap B_{3}\cap B_{4}) \ $$\rightarrow P(W_{1}).P(W_{2}/W_{1}).P(B_{3}/(W_{1}\cap W_{2}).P(B_{4}/(W_{1}\cap W_{2}\cap B_{1}))$
$\rightarrow (6/15).(5/14).(9/13).(8/12)$
$\rightarrow 6/91$
Or
simply using multiplication principle and drawing one ball at a time will give
the probability of drawing $1^{st}$ ball which is white is $6/15$
the probability of drawing $2^{nd}$ ball which is white is $5/14$
the probability of drawing $3^{rd}$ ball which is black is $9/13$
the probability of drawing $4^{th}$ ball which is black is $8/12$
$\rightarrow (6/15).(5/14).(9/13).(8/12)$
$\rightarrow 6/91$
Your answer should be $\binom{6}{2}*\binom{9}{2}/\binom{15}{2}*\binom{13}{2}$ $\because$ balls are to be randomly selected without replacement