Colored balls probability - color of remaining balls

Question

A bowl contains 5 white and 8 black balls. We proceed to pick out balls one by one (without putting any back in the bowl) until only same-color balls remain in. What should I think to calculate the probability of only white balls remaining in the bowl?

Answer 1

Write the arrangements of $BBBBBBBBWWWWW$ out such that each one ends with a $W$, and count these.

To avoid repetition, consider the endings of $BW$, $BWW$, $BWWW$, $BWWWW$, $BWWWWW$, and then arrange the remaining balls in front of them.

So in total:

$$\binom{11}{4}+\binom{10}{3}+\binom{9}{2}+\binom{8}{1}+\binom{7}{0}$$

$$=330+120+36+8+1$$

$$=495$$

The probability is then $\frac{495}{\binom{13}{5}}=\frac{5}{13}$

which is $\frac{5}{5+8}$ because the numerator is $\binom{12}{4}$ by the hockey-stick identity, and the denominator is $13\cdot\binom{12}{4}\cdot\frac15$.

Answer 2

To flesh out the hint in the comments:

Imagine that you draw out all the balls, even though they might be mono-chromatic well before you are done.

The probability that you want is just the probability that the last ball drawn is white so the answer is $$\boxed {\frac {5}{5+8}=\frac 5{13}\approx .384615}$$

In general, starting with $W$ white balls and $B$ black, the answer is $$\frac {W}{W+B}$$