Combination Probability

There are 10 students and 7 teachers. Four members of the committee are to be choosen from them randomly. What is the probability of the committee having more than 1 students?

I have been trying to use the n!/n!(n-r) formula, but I dont' know how to plug the things in properly. Please help!

Quote:

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Originally Posted by tttcomrader

There are 10 students and 7 teachers. Four members of the committee are to be choosen from them randomly. What is the probability of the committee having more than 1 students?

I have been trying to use the n!/n!(n-r) formula, but I dont' know how to plug the things in properly. Please help!

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I think you should divide this problem into several cases.

Exactly 1
C(10,1)*C(7,3)/C(17,4)

Exactly 2
C(10,2)*C(7,2)/C(17,4)

Exactly 3
C(10,3)*C(7,1)/C(17,4)

Exactly 4
C(10,4)*C(7,0)/C(17,4)

And add them up to get your answer.
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There is an easier way.
Since we want either 1,2,3, or all 4.
Find the probably of its negation, that is, 0.

Hence,
C(10,0)*C(7,4)/C(17,4)

Then 1 (total probability) subtract this result.
To get the same answer.