# probability problem counting techniques.

• Sep 26th 2007, 10:23 PM
rbenito
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.
• Sep 27th 2007, 12:58 AM
F.A.P
Quote:

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

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

or equivalently through

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

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 ...
• Sep 27th 2007, 04:29 AM
Soroban
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: . ${15\choose4} \:=\:1365$ possible committees.

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

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

• Sep 27th 2007, 04:42 AM
F.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.....)
• Sep 27th 2007, 07:50 AM
Soroban
Hello, F.A.P!

Quote: