<s>[INST] What is the probability that at least one of the committee members is in upper 6? One of the questions in my workbook read:

The school bowling club has $20$ members from upper 6 and $10$ from lower 6.The club wishes to form a committee consisting of $3$ members. Find the probability that at least one of the committee members is from upper 6.

My answer is $311/609$ while the book's answer is $197/203$. Please explain to me why I did not get the same answer as the book and how to get the correct answer.
My working:
At least one of the committee members is from upper 6 means either one of the committee members is from upper 6 or two of the committee members are from upper 6 or three of the committee members are from upper 6.
Probability that one of the committee members is from upper 6
$$= \frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28} = \frac{15}{203}$$
Probability that two of the committee members are from upper 6 $$= \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{10}{28}= \frac{95}{609}$$
Probability that three of the committee members is from upper 6
$$= \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{18}{28} = \frac{57}{203}$$
Probability that at least one of the committee members is from upper 6
$$=\frac{15}{203}+\frac{95}{609}+\frac{57}{203}=\frac{311}{609}$$
 [/INST] Your answer is not correct because your calculation does not reflect the fact that the committee selection is not ordered.  So for instance, $$\frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28}$$ represents the probability that the first member selected is from upper 6, then the following two members are selected from lower 6.  Since selected members do not have an ordering, the actual probability is $3$ times this value, since for any such selection, there are $3$ positions for the member that is from upper 6.
Similarly, the second probability you write needs to be multiplied by $3$.  But the last probability, in which all members are from upper 6, does not need to be multiplied by $3$ since all of them are from the same group.
Thus the corrected calculation would look like this:
$$3 \cdot \frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28} + 3 \cdot \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{10}{28} + \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{18}{28} = \frac{197}{203}.$$
However, none of this is necessary.  By far the easiest solution is to compute the complementary probability; i.e., the probability that none of the selected members are from upper $6$.  This occurs with probability
$$\frac{10}{30} \cdot \frac{9}{29} \cdot \frac{8}{28} = \frac{6}{203},$$ thus the desired probability that at least one member is from upper 6, is $$1 - \frac{6}{203} = \frac{197}{203}.$$
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