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? 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.
 A: 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 $\frac{406}{4060}= 0.1$
A: 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 $\frac{\text{no. of favourable ways}}{\text{total number of ways.}}$
A: 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.
A: After you've been chosen for a committee, there are $29 \choose 2$ ways of choosing the other two members of the committee.
A: 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.
A: 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.
