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A box contains 8 pieces each of milk chocolate, white chocolate, and dark chocolate. The box is passed around the six of us, with each person taking 4 pieces. Assume that each person chooses at random without replacement from the available pieces. I am the last person to whom the box is passed. Find the chance that I pick 4 dark chocolates.

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What have you tried, is this homework? –  Noturab Apr 17 '13 at 10:18
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3 Answers

The fact that there are 6 people picking is irrelevant. The fact that you pick last is irrelevant. The fact that there are milk and white is irrelevant.

You get 4 chocolates, for them to be all dark you have to choose 4 dark from your 4 choices.

$$p=\frac{8}{24}\frac{7}{23}\frac{6}{22}\frac{5}{21}$$

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HINT:

You have to calculate the number of combinations to pick 8 milk chocolates, 8 white chocolates and 4 dark chocolates among 3 persons.
So let $X$ be the number of combinations that 3 persons can pick from $8$ mc, $8$ wc and $4$ dc.
Let $Y$ be the number of combinations that 3 perons can pick from $8$ mc,$8$ wc and $8$ dc.
Than you will get your probability $$P= \frac{X}{Y}$$

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Here's a generalization:

There are $m$ sets size $r$ each, so the total number of items is $mr=n$. $k$ people select $s$ items each, hence $ks$ items are selected before the last person makes his selection. The first $k$ people can select anything, but leave at least $s$ items from the marked set $j$. If this is not the case, then the probability of your event is obviously 0.

The number of ways to select any $ks$ items out of $n-s$ is $\binom{n-s}{ks}$, the number of ways to select $ks$ items out of $n$ is $\binom{n}{s}$, the number of ways to select exactly $s$ items out of the marked set $j$ is $\binom{r}{s}$, and the number of ways to select $s$ items out of the remaining $n-ks$ is $\binom{n-ks}{s}$, hence the probability of your event is $$ P(S)=\frac{\binom{n-s}{ks} \cdot \binom{r}{s}}{\binom{n}{ks} \cdot \binom{n-ks}{s}} $$ This can be easily extended to the case when each person selects $t>s$ items.

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