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raleung2
Group Title
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!
 2 years ago
 2 years ago
raleung2 Group Title
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!
 2 years ago
 2 years ago

This Question is Closed

zepp Group TitleBest ResponseYou've already chosen the best response.1
dw:1335841523507:dw
 2 years ago

zepp Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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 \[\left({3 \over 20}\cdot{2 \over 19}\cdot{1 \over 18}\right)+\left({3 \over 20}\cdot{17 \over 19}\cdot{2 \over 18}\right)+\left({17 \over 20}\cdot{3 \over 19}\cdot{2 \over 18}\right)+\left({17 \over 20}\cdot{16 \over 19}\cdot{3 \over 18}\right)\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
If you just want the probability of all three places being claimed by a female horse, then it's just \[{3 \over 20}\cdot{2 \over 19}\cdot{1 \over 18}\]
 2 years ago

zepp Group TitleBest ResponseYou've already chosen the best response.1
Oh I see, thanks for the explanation :D
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You're very welcome.
 2 years ago
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