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Math Help - probability of choosing 4 person

  1. #1
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    probability of choosing 4 person

    for part b , my working is (12c2 x 18c2)/(35c4) = 0.192


    thers're 12 women doctor , 2 are chosen , there're 18 male left, 2 are chosen . no condition= choose 4 person from 35 person. but the ans is 0.807
    probability of choosing 4 person-img_20140725_054858-2-.jpg
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  2. #2
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    Re: probability of choosing 4 person

    How many possible panels have two women and two men? That would be {_{17}C_2}\cdot {_{18}C_2}. So, your denominator is wrong.

    Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49 I dunno where 0.807 comes from.
    Last edited by SlipEternal; July 25th 2014 at 02:12 AM.
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    Re: probability of choosing 4 person

    Quote Originally Posted by SlipEternal View Post
    How many possible panels have two women and two men? That would be {_{17}C_2}\cdot {_{18}C_2}. So, your denominator is wrong.

    Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49 I dunno where 0.807 comes from.

    the question ask for 4 person to be chosem in the panel disussion am i right? which include 2 woman doctor..
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    Re: probability of choosing 4 person

    Quote Originally Posted by SlipEternal View Post
    How many possible panels have two women and two men? That would be {_{17}C_2}\cdot {_{18}C_2}. So, your denominator is wrong.

    Anyway, it is just asking about the women. The men don't factor into it at all. Given that you choose two women at random, what is the probability that they are both doctors. That would be \dfrac{12\cdot 11}{17\cdot 16}\approx 0.49 I dunno where 0.807 comes from.
    the question ask for 4 person to be chosem in the panel disussion am i right? which include 2 woman doctor..
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    Re: probability of choosing 4 person

    The problem said assuming there are exactly two women on the panel, what is the probability that both are women. This is exactly the same as asking "If two women are selected from a group of 17, 12 of whom are doctors, what is the probability they are both doctors. The number of men is not relevant.
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  6. #6
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    Re: probability of choosing 4 person

    Had the question been worded, "If at least two women are chosen for the panel, what is the probability that at least two women doctors are chosen", then the answer would be:

    \dfrac{ {_{12}C_2}\cdot {_{23}C_2} + {_{12}C_3}\cdot {_{23}C_1} + {_{12}C_4}\cdot {_{23}C_0} }{ {_{17}C_2}\cdot {_{18}C_2} + {_{17}C_3}\cdot {_{18}C_1} + {_{17}C_4}\cdot {_{18}C_0} } = \dfrac{1309}{2084} \approx 0.628

    I don't think I know of a question similar to the one asked that could yield a probability as high as they want.
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