I already figured out the number of possible combinations that 3 people can be chosen for the committee which is $30\choose 3$ $= 4060$. Just don't know how to find the next step.
2026-04-01 23:35:00.1775086500
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A class of 30 students must form a committee of three. You are in the class, what is the probability that you will be chosen at random for committee?
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After you've been chosen for a committee, there are $29 \choose 2$ ways of choosing the other two members of the committee.
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There are $\binom {30}3$ possible committees and there are $\binom {29}3$ possible committees you are not on. From this - don't expand the terms, as a lot will cancel, and this will save you lots of time - you can find the probability you are not on the committee, and thence the probability that you are.
Alternatively think of the committee places as the first three in a row of $30$ - you are equally likely to be in each place. And you should be able to compute the probability you are in one of the first three places from that.
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How many of those $\begin{pmatrix}30 \\ 3\end{pmatrix}= 4060$ possible committees include you? Assuming you are chosen, that leaves 29 people from whom to select two. The number of ways you can do that, using that same formula is $\begin{pmatrix}29 \\ 2\end{pmatrix}= 406$. The probability is $\frac{406}{4060}= 0.1$
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If you select 3 people out of thirty, the probability that any one person has been selected is equal to the fraction of people that have been selected. In this case, 10% of people are on the committee, so the probability that any given person is on the committee is 10%.
Worrying too much about the number of possible committees in this case seems like overkill.
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You can separate this problem into 3 independent random selection events where the probability of you being chosen is 1 / the number of the people in the selection pool. The probability of any one person being chosen in each one of these random selections is 1/30, 1/29, and 1/28 (assuming the same person cannot be selected twice).
The probability of you being selected in one of those 3 events is the same as 1 - the probability that you are not selected in any of them. Therefore, we can represent the problem as so:
1 - ((29/30) * (28/29) * (27/28)) = 0.1
In this problem, any one person has a 10% chance of being selected to be in the committee, including yourself.
If you have to be chosen, then the only thing to decide is who the other two are going to be. So just make a two person committee and add yourself to it. How many ways to make a two person committee? After that, probability is just $\frac{\text{no. of favourable ways}}{\text{total number of ways.}}$