An urn initially contains 6 white and 8 black balls. Each time a ball is selected, its color
is noted. If the selected ball is white, then it is replaced in the urn along with 3 other black balls. If the
selected ball is instead black, then it is replaced along with 2 other white balls. Compute the probability
that the first 2 balls selected are black and the third selected ball is white.

My answer:

$P(B\_1B\_2W\_3)=\frac{{8 \choose 1} {6 \choose 0}}{{14 \choose 1}}\frac{{8 \choose 1} {8 \choose 0}}{{16 \choose 1}}\frac{{8 \choose 0} {10 \choose 1}}{{18 \choose 1}}$

which I think equals 10/63

Is my understanding of this problem correct?