1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability of invites homework

  1. Dec 4, 2007 #1
    Sorry about not having the right writing way on the definitions, I'll update that soon.

    1. The problem statement, all variables and given/known data

    A woman takes a.. drawing among her 11 friends(ie 0 to 11 might arivve. Also 6 are women and 5 are men). What is the probability that by conincidence is as much women as men among the company?

    2. Relevant equations

    binominalkoefficient(I think thats relevant)
    (p)
    (r)
    (p over r)

    .. and then I don't know.

    3. The attempt at a solution

    I tried (((6 over 0)*(5 over 1))\(11 over 1)((6 over 1)*(5 over 2))\(11 over 3) ((6 over 2)*(5 over 3))\(11 over 5)((6 over 3)*(5 over 4))\(11 over 7)((6 over 4)*(5 over 5))\(11 over 9)) / 4. But that's something like 0.316. And the is 0.164.


    Problem 2. I have some minor problems with Bayes sentence.

    eg. 0.6 of women have aids, 0.4 of them testes positive. 0.4 have not aids, and .03 of them also get positive.

    The bottom of the sentence is ok, just P(B), But when do you know it's 0.6 * P(B|A), or 0.6* (just)0.4, or just 0.6. I mess up there. (eg if you could explain it throu how the different questions relating to those differences would look)
     
  2. jcsd
  3. Dec 4, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you saying the woman invites 11 friends, each of whom may or may not come? Seems to me you would need to know the probability each will or will not come.

    "I tried (((6 over 0)*(5 over 1))\(11 over 1)((6 over 1)*(5 over 2))\(11 over 3) ((6 over 2)*(5 over 3))\(11 over 5)((6 over 3)*(5 over 4))\(11 over 7)((6 over 4)*(5 over 5))\(11 over 9)) / 4. But that's something like 0.316. And the [answer?] is 0.164."
    It would help if you said WHY you tried that!
     
  4. Dec 4, 2007 #3
    Yeah, you could say it like that.

    I tried that answer because as you said that would be the probability for each combination. Then I added each(didn't know if to add or multiply, but neither work anyway), .. and then I guess just by intuition I divide by 4 as to get the 'line' that gives the general outcome(some would give higher probability, some lower)(don't know quite how to spell this in english, sorry!!)
    It doesn't feel right, but I didn't know what else to do and it was my first thought.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Probability of invites homework
Loading...