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Paul opened a bag of candy and found the following colors of candy: $6$ red, $5$ blue, $12$ green, $9$ yellow, and $13$ orange. What is the probability he will choose two candies that are not yellow?

The question doesn't state whether Paul places the candy back in the bag after he has picked at least one, so can I assume the question is asking for $2$ not yellow without replacement? On my sheet the answer line looks like this:

P(2 not yellow)

I got the probability that Paul doesn't choose yellow ($36/45$) but not sure if it is right.

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2 Answers

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Since Paul is presumably interested in eating the candies he selects, it is reasonable to assume that the candies are chosen without replacement. There is also an implicit assumption that the candies are selected without looking in the bag, which ensures that the selection is random.

You made the assumption that the selections are made successively, which is not necessary. We consider two approaches:

1. The candies are selected randomly in succession without replacement.
2. Two candies are selected randomly at once.

The candies are selected randomly in succession without replacement:

There are $6 + 5 + 12 + 9 + 13 = 45$ candies in the bag, of which $9$ are yellow and $45 - 9 = 36$ are not yellow. Thus, on the first draw, the probability that the first candy is not yellow is $36/45$. If the first candy is not yellow, that leaves $44$ candies in the bag, of which $35$ are not yellow. Thus, the probability that the second candy is also not yellow is $35/44$. Hence, the probability that both candies are not yellow is $$\frac{36}{45} \cdot \frac{35}{44} = \frac{4}{5} \cdot \frac{35}{44} = \frac{7}{11}$$

Two candies are selected randomly at once:

There are $\binom{45}{2}$ ways to select a subset of two of the $45$ candies and $\binom{36}{2}$ ways to select a subset of two of the $36$ candies that are not yellow. Hence, the probability that the two candies selected are not yellow is $$\frac{\dbinom{36}{2}}{\dbinom{45}{2}} = \frac{\dfrac{36!}{2!34!}}{\dfrac{45!}{2!43!}} = \frac{\dfrac{36 \cdot 35 \cdot 34!}{2 \cdot 1 \cdot 34!}}{\dfrac{45 \cdot 44 \cdot 43!}{2 \cdot 1 \cdot 43!}} = \frac{\dfrac{36 \cdot 35}{2}}{\dfrac{45 \cdot 44}{2}} = \frac{36 \cdot 35}{2} \cdot \frac{2}{45 \cdot 44} = \frac{36 \cdot 35}{45 \cdot 44} = \frac{7}{11}$$

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The answer is approximately 0.63

(36/45)*(35/44) = 0.63

The first probability is with all 36 options. Then the next probability is with 44.

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