Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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
  1. raleung2 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    help!

    • 2 years ago
  2. zepp Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1335841523507:dw|

    • 2 years ago
  3. zepp Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  4. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  5. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 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
  6. zepp Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh I see, thanks for the explanation :D

    • 2 years ago
  7. KingGeorge Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You're very welcome.

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.