can anyone helpe with this problem?:

Companies A,B,C,D, and E each send three delegates to a conference. A committee of 4 delegates, selected by lot, is formed. Determine the probability that company A has exactly one representative on the commitee.

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- September 26th 2007, 10:23 PMrbenitoprobability problem counting techniques.
can anyone helpe with this problem?:

Companies A,B,C,D, and E each send three delegates to a conference. A committee of 4 delegates, selected by lot, is formed. Determine the probability that company A has exactly one representative on the commitee. - September 27th 2007, 12:58 AMF.A.P
There will be 3 A-delegates and 12 non-A-delegates. Probability of choosing exactly one A-delegate is found through

or equivalently through

,

where we have to multiply by 4 at the end because the A-delegate can be either the first, second, third or last person to be chosen ... - September 27th 2007, 04:29 AMSoroban
Hello, rbenito!

Quote:

Companies A, B, C, D, and E each send three delegates to a conference.

A committee of 4 delegates, selected by lot, is formed.

Determine the probability that company A has exactly one representative on the commitee.

There are 15 delegates; 4 are chosen for the committee.

. . There are: . possible committees.

There are 3 delegates from company A and 12 Others.

To pick one A and 3 Others, there are: . ways.

Therefore, the probability is: .

- September 27th 2007, 04:42 AMF.A.P
You're right Soroban... I read the question too fast and had 4's stuck in my head...:D there where only three from each company... I've now changed the numbers in my previous post... (where an alternative approach also is presented.....)

- September 27th 2007, 07:50 AMSoroban
Hello, F.A.P!

Quote:

I read the question too fast and had 4's stuck in my head.

(If I had a nickel for every time I've done that . . . )

And I like your alternate solution.

It's very neat . . .__if__everyone understands the "times four".