Probability of girls vs. boys

lp107

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Apr 1, 2008
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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

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Jan 29, 2005
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At least one is opposite of none.
\(\displaystyle 1 - \frac {\binom {8}{4}} {\binom {20}{4}}.\)
 

soroban

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Jan 28, 2005
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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?

\(\displaystyle \text{There are: }\;{20\choose4} \:=\:4845\text{ possible choices.}\)


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


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

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

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

 
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