7 white balls and 5 black balls. Find the probability that the 6th ball drawn is white, while...

Question

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!?

Answer 1

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 $\frac{25}{132}$.

Answer 2

The probability that, in the first five balls, exactly three are black, is $$\frac{\binom{5}{3}\binom{7}{2}}{\binom{12}{5}} = \frac{35}{132}$$ since there are $\binom{5}{3}$ ways of drawing the black balls, $\binom{7}{2}$ ways of drawing the white balls, and $\binom{12}{5}$ ways of drawing five balls without restriction.

After those five are drawn, there are five white and two black balls left, so the probability that the sixth ball is white is $\frac{5}{7}$. Multiplying those together gives $\frac{25}{132}$.

Answer 3

Alternatively: The outcomes are: $$BBBWWW, BBWBWW, BBWWBW, BWBBWW, BWBWBW, \\ BWWBBW, WBBBWW, WBBWBW, WBWBBW, WWBBBW.$$ Hence the probability is: $$10\cdot \frac{5\cdot 4\cdot 3\cdot 7\cdot 6\cdot 5}{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7}=\frac{25}{132}.$$

Answer 4

As posed:

No combinatorics necessary, it's an obfuscation.

Because exactly three black balls (and therefor two white) were drawn first, when you go to draw the sixth ball there must be 2 black and 5 white remaining.

Thus, the probability of drawing white is simply 5/7.

...but as @jpmc26 points out, the question probably means to ask, "what is the probability of drawing three blacks and two whites, and then drawing a white."