[lyogi123]

There are 30 people in a class. We want to assign them to three committees-- the environmental committee (5 people), student government (6 people) and teacher assistants (7 people).
If the committees are chosen randomly, what is the chance that a given person winds up on some committee?
(For this I was thinking C(30,1) is the possibility that we choose a certain person and C(30,18) is the ways of choosing all the committees. So maybe C(30,1)/C(30,18)? I'm a little confused on this whole thing... )

If a person does wind up on a committee, what is the chance that it is the environmental committee?


Suppose people can be on more than one committee. If the committees are chosen randomly what is the chance that a person winds up on more than one committee?

any help/explanation would be greatly appreciated! I'm trying to go over all the practice problems, so that I can understand the concepts as best as possible before I take my test! Thank you in advance for any explanations or help u can provide

[ebaines]

You are over-thinking these - these questions are easier than you realize! For each simply remember that the odds of something occurring equals the number of of possible "winners" divided by the number of possible outcomes. For example, for the first problem there are 18 people chosen to be on a committee (I assume no person can be on two committees for this part) out of 30. So - what are the odds of any person being on a committee? For the second, there are 18 people on committees, of whom 5 are on the environmental committee - what are the odds that a person who is in a committee is on the environmental committee?

For the last part - the odds of being on more than one committee is equal to one minus the sum of odds of being on no committee or being on exactly one committee. The second part of that calculation is the only case in any of these problems where you need to use combinations.