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.
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 ...
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: .$\displaystyle \frac{660}{1365} \;=\;\frac{44}{91}$
Hello, F.A.P!
I thought that's what happened.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".