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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?

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    $\begingroup$ You are correct. Nice work. $\endgroup$ – N. F. Taussig Feb 16 '15 at 1:28
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Yes your method works but you have avoided the KISS principle.

Start with:

$$P(B_1)=\frac{8}{14}$$

return the black ball and add two white balls to give:

$$P(B_2|B_1)=\frac{8}{16}$$

return the black ball and add two white balls to give:

$$P(W_3|B_1B_2)=\frac{10}{18}$$

$$\begin{align} P(B_1B_2W_3)&=P(W_3|B_1B_2)P(B_2|B_1)P(B_1)\\ &=\frac{10}{18}\frac{8}{16}\frac{8}{14}\\ &=\frac{10}{63} \end{align}$$

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