1. probability 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.

2. Originally Posted by rbenito
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.
There will be 3 A-delegates and 12 non-A-delegates. Probability of choosing exactly one A-delegate is found through

$\displaystyle \frac{{3\choose 1}{12\choose 3}}{{15\choose 4}}=44/91$

or equivalently through

$\displaystyle \frac{3}{15}\cdot \frac{12}{14}\cdot \frac{11}{13}\cdot \frac{10}{12}\cdot 4=44/91$,

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 ...

3. Hello, rbenito!

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: .$\displaystyle {15\choose4} \:=\:1365$ possible committees.

There are 3 delegates from company A and 12 Others.
To pick one A and 3 Others, there are: .$\displaystyle {3\choose1}\cdot{12\choose3} \:=\:660$ ways.

Therefore, the probability is: .$\displaystyle \frac{660}{1365} \;=\;\frac{44}{91}$

4. You're right Soroban... I read the question too fast and had 4's stuck in my head... there where only three from each company... I've now changed the numbers in my previous post... (where an alternative approach also is presented.....)

5. Hello, F.A.P!