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lp107
04-01-2008, 11:59 PM
Hi, I was given practice problems for our exam, and there is this one question that I can not get. The question is: There are 20 people in a class, 8 are men, 12 are women, in a random set of 4, what is the probability of getting just women, and what is the probability of getting 1 man and 3 women? I tried doing it several different ways, but it never comes out to the answer my professor gave me!

Thanks for helping!
Lauren

pka
04-02-2008, 07:53 AM
At least one is opposite of none.
1 - \frac {\binom {8}{4}} {\binom {20}{4}}.

soroban
04-02-2008, 08:39 AM
Hello, Lauren!


There are 20 people in a class, 8 are men, 12 are women.
Four a chosen at random.

(a) What is the probability of getting just women?
(b) What is the probability of getting 1 man and 3 women?
\text{There are: }\;{20\choose4} \:=\:4845\text{ possible choices.}


\text{(a) There are: }\:{12\choose4} \:=\:495\text{ ways to choose 4 women.}
. . \text{Therefore: }\;P(\text{4 women}) \:=\:\frac{495}{4845} \;=\;\frac{33}{323}


\text{(b) There are: }\;{8\choose1} \:=\:8\text{ ways to choose 1 man.}
\text{There are: }\;{12\choose3} \:=\:220\text{ ways to choose 3 women.}

\text{Hence, there are: }\:8\times220 \:=\:1760\text{ ways to choose 1 man and 3 women.}

. . \text{Therefore: }\;P(\text{1 man, 3 women}) \;=\;\frac{1760}{4845} \;=\;\frac{352}{969}