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A convenience store has exactly three packages of plain M&M's, two packages of peanut M&M's, one package of dark chocolate M&M's, one package of peanut butter M&M's, and one package of almond M&M's. Select two packages of candy. Calculate the probability you select exactly one package of plain M&M's, and the other package is not plain.

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We assume that the selection is done "at random," with every pair of packages equally likely to be chosen. This is an unrealistic assumption, since presumably most people have preferences. However, that is the assumption that the problem-setter expects us to make.

There are $8$ packages, of which $3$ are plain. We can choose $2$ packages in $\binom{8}{2}$ ways, all equally likely.

Now we count the number of ways to choose one package of plain, and one package of not plain. The plain package can be chosen in $\binom{3}{1}$ ways. For every such choice, the not plain package can be chosen in $\binom{5}{1}$ ways, for a total of $\binom{3}{1}\binom{5}{1}$ ways. So the required probability is $$\frac{\binom{3}{1}\binom{5}{1}}{\binom{8}{2}}.$$ Calculate: our probability is $\dfrac{15}{28}$.

Another way: Imagine instead picking the packages one at a time. Let P stand for "plain" and N for "not plain." Then there are two "patterns" that give us altogether one plain and one not plain, namely PN and NP.

We calculate the probability of the pattern PN. The probability our first package is plain is $\dfrac{3}{8}$. Given that we got a plain on the first pick, there are $7$ packages left, so the probability of not plain on the second pick is $\dfrac{5}{8}$. Thus the probability of the pattern PN is $\dfrac{3}{8}\cdot\dfrac{5}{7}$.

A similar calculation shows that the probability of NP is $\dfrac{5}{8}\cdot \dfrac{3}{7}$. (This probability actually must be the same as the probability of PN, so the second calculation was not necessary.) We conclude that the required probability is $(2)\left(\dfrac{15}{56}\right)$, or equivalently $\dfrac{15}{28}$.

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Since all the answer asks for is plain M&Ms vs not plain M&Ms, we can simplify the information to 3 packages of plain and 5 packages of not plain.

Now the number of ways we can choose 1 plain package from 3 possibilities is $3\choose1$ and the number of ways to choose 1 not plain package from 5 possibilities is $5\choose1$.

From this, we find that the number of ways to choose one plain package and one not plain package is $${3\choose1} {5\choose1} = 15$$

Now, the total number of ways to choose 2 packages from 8 possibilities is $${8\choose2} = 28$$ so the probability of picking the desired set of packages is $\frac{15}{28}$.

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How do you know when you should treat a subset as indistinguishable in a sample space? In other words, in this case you have 3 plain packages: $P_1, P_2, P_3$ but I would consider 1 plain and 1 peanut M&M the same. But the way you do it you distinguish between $(Plain_1, Peanut_1) and (Plain_2, Peanut_2)$. – user1527227 May 2 '13 at 15:48

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