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Homework Help: Help with a simple probability problem

  1. Oct 30, 2006 #1
    Ok, this seems like an easy enough problem, but I'm stumped...

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    If a motorcycle driver is selected at random from the US population, the probability of that driver being a male is 0.95
    If 20 motorcycle drivers are selected at random from the population, what is the probability that exactly 8 will be women? What is the probability that at least 8 will be women?
    ----------------------------------------------------------------------

    I've got that the chance of selecting a female is 1/20. And because the population is so large I can treat the events as independent.

    I can easily find the probability of selecting 8 women in a row. What's throwing me off is the fact that there are 20 selections, so I know that the probability of selecting 8 total isn't going to be the same as selecting 8 in a row. I can't figure out how to account for the fact that 20 selections increases the probability of 8 women being selected.
    I'm also thrown off by the at least 8 vs. exactly 8 part of the problem.

    Any hints would be great. Thanks! :biggrin:
    -GeoMike-
     
    Last edited: Oct 30, 2006
  2. jcsd
  3. Oct 30, 2006 #2

    Office_Shredder

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    if event A has a probability p of occuring, and an event B has a probability 1-p of occuring, and one event or the other occurs n times, you can characterize the probability by

    pk(1-p)n-k

    where k is the number of times A happens (this assumes order counts. In your problem it doesn't, so you'll need to fix it for that)
     
  4. Oct 31, 2006 #3
    Ok, that makes sense. So, the probability of getting 8 females and 12 males (in a specific combination) would be:

    (1/20)8(19/20)12

    To account for the fact that any arrangement of 8 females and 12 males is acceptable, I'd just multiply that probability by the total number of combinations:

    (20!/(12!*8!)) * (1/20)8(19/20)12 = P(exactly 8 women)

    Right?

    And to find P(at least 8 women) I'd just sum the probabilities of exactly 8 through exactly 20 women:

    [tex]\sum_{i=8}^{20}{\frac{20!}{(20-i)!i!}(1/20)^{x}(19/20)^{20-x}}[/tex]

    Right?

    Thanks,
    -GeoMike-
     
  5. Oct 31, 2006 #4
    of course, you'll have fewer things to sum of you do 0 to 7 females... and something about 1-p

    Although, in this case, it's not that big a deal. But, if it was "at least 2 females", it'd be far quicker to find the probabilities of 0 females and 1 female, then subtracting from 1.
     
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