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Let's say it's me and 2 other friends are at a raffle. There are 10 total participants. 3 of the 10 of us will win something, I don't care about what we win. A person can't win more than once. What is the probability that exactly two of us win something?

I'm thinking that the total outcomes is ${10 \choose 3}$ which is 120. But then the numerator I'm struggling. Would it be ${3 \choose 2} * 7$? If so, I'm confusing myself over it, so can someone please explain why this would/wouldn't work?

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2 Answers 2

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This maybe the answer you seek. This result yields the probability of none of your group are winners. $$ \frac{\binom{3}{0} \binom{7}{3}}{\binom{10}{3}} $$

This is the probability of one winning in your group, I think you can figure out the rest.

$$ \frac{\binom{3}{1} \binom{7}{2}}{\binom{10}{3}} $$

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Numerator: 3C2 Denominator: 10C2

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  • $\begingroup$ Can you explain how you got it? $\endgroup$
    – rb612
    Feb 6, 2016 at 7:25
  • $\begingroup$ This is being automatically flagged as low quality, probably because it is so short. At present it is more of a comment than an answer by our standards. Can you expand on it? We can also turn it into a comment. $\endgroup$ Feb 6, 2016 at 7:52
  • $\begingroup$ So nos of ways any 2 contestants can win is 10C2, which will be denominator. $\endgroup$
    – tooshrit
    Feb 6, 2016 at 8:06
  • $\begingroup$ And you are 3 contestants and any 2 of them wining can happen in 3C2 ways, which will be numerator. $\endgroup$
    – tooshrit
    Feb 6, 2016 at 8:07
  • $\begingroup$ @tooshrit - but I think you have to factor in that you're choosing 3, not 2 people in the denominator. $\endgroup$
    – rb612
    Feb 6, 2016 at 23:12

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