Solution
= Solution
The <probability> that the first ball is white is $x$. Partitioning that event according to the second colour shows that $P(WB)=x-p_w$. Uniform <sampling without replacement> makes the two draw positions exchangeable, so $P(BW)=P(WB)$. Therefore
$$
\boxed{p_d=2(x-p_w).}
$$
For a direct count, the same value is $2(xN)((1-x)N)/[N(N-1)]$.