# Compound Probability and Counting Principle

This very straight forward probability question is somehow confusing to me. The box contains $$3$$ maple donuts, $$1$$ choc donut, and $$2$$ glazed. One donut is chosen randomly without replacement and then a second is chosen.

Q1: Create an organized list to represent the sample space for this situation.

In this situation I would think that the order of choice does not matter as you come away with two donuts and the order in which they were chosen does not matter. So, I get $$15$$ items in the sample space. This is the same number I would get if I were to use the mathematical combination computation of $$6$$ items taken $$2$$ at a time. In other words unordered sampling without replacement. My tree matches with the combination of $$6$$ taken $$2$$ at a time. $$15$$ possible ways.

Note: I assume that the $$3$$ maple donuts are indistinguishable; same with $$2$$ glazed.

I get stuck when the problem then says, "Use the counting principle to find how many different ways the two donuts can be chosen."

This would seem to indicate that since the first donut can be chosen $$6$$ different ways and the second donut $$5$$ different ways I would get $$30$$. This I know is the answer if I were to chose $$2$$ donuts from $$6$$ where order matters (permutation). However, it would seem to me that order does NOT matter in this case.

How can I use the fundamental counting principle in this case to get $$15$$ ways? Or is the real answer $$30$$?

suggests that the order in which the donuts are chosen matters. Therefore, your sample space consists of ordered pairs of the form (first donut selected, second donut selected). As you observed, there are six possible selections for the first donut and five possible selections for the second donut, giving $$6 \cdot 5 = 30$$ outcomes in your sample space.
If you were interested in the probability that both selected donuts were maple donuts, you would have $$3 \cdot 2 = 6$$ favorable cases, giving $$\Pr(\text{both maple}) = \frac{3 \cdot 2}{6 \cdot 5} = \frac{1}{5}$$ That said, the same answer is obtained if the order in which the donuts are selected does not matter. As you noted, there are $$\binom{6}{2}$$ ways to select a subset of two of the six donuts. Of these, $$\binom{3}{2}$$ such choices result in the selection of two maple donuts. Hence, $$\Pr(\text{both maple}) = \frac{\dbinom{3}{2}}{\dbinom{6}{2}} = \frac{1}{5}$$ You will obtain the same probabilities regardless of whether your sample space consists of ordered pairs or two-element subsets, but the way the first question is worded and the reference to the Fundamental Counting Principle in the second question suggests that the order in which the donuts are selected matters, which is why I took the sample space to be the set of ordered pairs.