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Probability: Choosing a girl from a group

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data
    You walk into your class the first day of classes, and you notice that
    there are 30 men and 20 women in the class already. Let's suppose you decide to choose
    two people from the class to be your study partners.

    If you choose your study partners at random, and given that at least one of your
    study partners is a woman, what is the probability of the event E that both of them
    will be women?
    A. 0.3167
    B. 1.9%
    C. 0.2405
    D. 0.1901

    2. Relevant equations
    In my Solution


    3. The attempt at a solution
    This seems like a simple problem but I cannot seem to get the numbers available as choices.
    My logic is is, W represents the event that you have picked a woman, and E represents that both of your partners will be women then.
    [tex]P(E|W)=\frac{P(E \cap W)}{P(W)}[/tex]
    So the numerator can simplify to..
    [tex]P(E|W)=\frac{P(E)}{P(W)}[/tex]
    This is because if E occurs, then W must have occured.
    So..
    [tex]P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]
    and
    [tex]P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}[/tex]

    But this doesn't work because the ratio of these two (From the formula) gives a number larger than one. Where am I going wrong? Do I use Bayes theorem?
     
  2. jcsd
  3. Feb 27, 2012 #2

    Dick

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    If P(W) is the probability of choosing at least one woman, there are two ways to do that, you could choose 1 woman and 1 man, or 2 women.
     
  4. Feb 27, 2012 #3
    Ohh! So it would be..
    [tex]P(W)=P(W|W)P(W)+P(W|M)P(M)[/tex]
    ??
    I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.
     
  5. Feb 27, 2012 #4

    Dick

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    I wouldn't write it that way. Just count out the cases using combinations like you are already doing. How may ways to do each?
     
  6. Feb 27, 2012 #5
    So choosing 1 woman and 1 man would be
    [tex]\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}[/tex]
    And choosing 2 women would be..

    [tex]\frac{\binom{20}{2}}{\binom{50}{2}}[/tex]

    Plugging this in gives me 0.2405, or answer C! Thanks! :D

    EDIT: Assuming that's the correct answer.....
     
  7. Feb 27, 2012 #6

    Dick

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    Well, that's what I get.
     
  8. Feb 27, 2012 #7

    Ray Vickson

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    I W = number of women you choose, you have been asked to find the conditional probability [itex]P\{W=2|W\geq 1\}. [/itex] We have [tex]P\{W=2|W\geq 1\} = \frac{P\{W = 2 \cap W \geq 1 \}}{P\{W \geq 1\}} = \frac{P\{W = 2\}}{1-P\{W=0\}}.[/tex] The numerator and denominator are easily comutable using the hypergeometric distribution. The numerator is [itex] {20 \choose 2}/{50 \choose 2},[/itex] while the denominator is [itex] 1 - {30 \choose 2}/{50 \choose 2}.[/itex]

    RGV
     
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