A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly, one at a time without replacement, until the box is empty. Find the probability that the 6th ball drawn is white, while before that exactly 3 black balls are drawn.
I thought the problem this way:
There are $\dfrac{12!}{7!5!}=792$ ways to order the balls in general. Then, if we fix black,black,black,white we have eight other balls to place in the other bins, which is $\dfrac{8!}{6!2!}=28$. Therefore the probability is $\dfrac{28}{792}=\dfrac{7}{198}$. But the book says the answer is $\dfrac{25}{132}$, why tho!?
Best Answer
First, we need the probability that among the first five balls, excactly three were black (and therefore two were white). This is a standard hypergeometric probability exercise, and the answer is $$ \frac{\binom53\binom72}{\binom{12}5} = \frac{35}{132} $$ Then, the probability that after this, the sixth ball is white. At this point there are seven balls left, and five of them are white. So the probability of drawing one of them as the seventh ball is $$ \frac{5}{7} $$ Now just multiply these two probabilities together, and you get your $\frac{25}{132}$.