4
$\begingroup$

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!?

$\endgroup$
1
  • 3
    $\begingroup$ I think you are wrong in assuming that the third, fourth and fifth balls drawn have to be black. I take the question to mean that precisely three of the first five balls drawn must be black. $\endgroup$ – Moriarty Feb 15 '18 at 22:51
16
$\begingroup$

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 $\frac{25}{132}$.

$\endgroup$
5
$\begingroup$

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 $\frac{25}{132}$.

$\endgroup$
2
  • $\begingroup$ Something still doesn't make sense to me... wouldn't $\binom{5}{3}\binom{7}{2}$ be counting black,black,black,white,white and ignoring all the other possible bbbww permutations? $\endgroup$ – the_dummy Feb 16 '18 at 2:26
  • $\begingroup$ @the_dummy It does and should: both the numerator and denominator are a number of combinations, not permutations. Including permutations in the numerator but not in the denominator will give an incorrect answer. You can work with permutations if you want, but you'll have to use them consistently. $\endgroup$ – hvd Feb 16 '18 at 7:04
4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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."

$\endgroup$
5
  • $\begingroup$ This ignore the probability of drawing exactly 3 black balls before it. $\endgroup$ – jpmc26 Feb 16 '18 at 3:15
  • 1
    $\begingroup$ @jpmc26 I read the probability of drawing exactly three black balls before it as 1. Do you think it's looking for white given three black, or white and three black? $\endgroup$ – fectin Feb 16 '18 at 3:17
  • 1
    $\begingroup$ The book's answer seems to imply the latter, and it seems unlikely that the problem statement would bother including irrelevant information. I agree the problem statement is somewhat ambiguous. $\endgroup$ – jpmc26 Feb 16 '18 at 3:20
  • 1
    $\begingroup$ @jpmc26 good point. I did wonder why everyone was trying to solve this the hard way... $\endgroup$ – fectin Feb 16 '18 at 3:22
  • $\begingroup$ I had the same interpretation, that it was asking for the probability of drawing white on the 6th draw given that exactly 3 previous draws were black. I assumed the point of the exercise was to recognize that the other information was irrelevant. Spotting the relevant information in probability calculations is an important skill, after all, and I've seen many problems where irrelevant information was intentionally given. $\endgroup$ – Paul Sinclair Feb 16 '18 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.