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In the 131st running of the Kentucky Derby in 2005, there were 20 contenders. If all 20 horses have an equal chance of winning and 3 of the horses are female, what is the probability that a female horse places first, second, and third? WORK PLEASE!

Mathematics
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help!
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Uh.. I think that 1st: 3/20 (3 chances on 20) 2nd: 3/19 (3 chances on 19, since a male horse already won) 3rd: 3/18 (3 chances on 18, since 2 male horse already won) I'm really not sure about my work, I haven't done any probability question for almost 2 years , so I'm a little bit rusty :s @KingGeorge

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Other answers:

For a female horse to place first, the chance is indeed 3/20. For a female horse to place second, you have to add the probability that a female horse placed first and second, and the probability that a female horse only placed 2nd. That would be \[{3\over20} \cdot {2\over19} + {17\over20}\cdot{3\over19}\] For a female horse placing third we need to consider all 4 possibilities of FFF, FMF, MFF, MMF. This results in a probability of \[\left({3 \over 20}\cdot{2 \over 19}\cdot{1 \over 18}\right)+\left({3 \over 20}\cdot{17 \over 19}\cdot{2 \over 18}\right)+\left({17 \over 20}\cdot{3 \over 19}\cdot{2 \over 18}\right)+\left({17 \over 20}\cdot{16 \over 19}\cdot{3 \over 18}\right)\]
If you just want the probability of all three places being claimed by a female horse, then it's just \[{3 \over 20}\cdot{2 \over 19}\cdot{1 \over 18}\]
Oh I see, thanks for the explanation :D
You're very welcome.

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