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I've been trying to work out the problem: If three people are selected randomly from a team of 7, what's the chance the captain is chosen? When I did it by drawing out a tree diagram I got 3/7 which seems like it should be right, but there are two problems. Firstly, when I do it another way, by calculating 6 choose 2 divided by 7 choose 3, I get a different answer. Secondly, both these answers seems to big, because if I then ask, what's the probability of choosing the captain or the vice, it should be two times that number, but that result is greater than 1.

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  • $\begingroup$ Something seems wrong since $\binom{6}{2}/\binom{7}{3}$ does equal 3/7. Furthermore twice 3/7 is not bigger than 1. Do note that twice 3/7 is not the correct value of the probability that the captain or the vice is chosen, since the event "captain is chosen" and the event "vice is chosen" are not mutually exclusive. $\endgroup$ – Will Orrick Sep 21 '13 at 22:51
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$$\frac{\binom{6}{2}}{\binom{7}{3}} = \frac{15}{35} = \frac37$$

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Indeed the probability of choosing the captain is $$\frac{\binom{6}{2}}{\binom{7}{3}}=\frac{3}{7}$$ by a counting argument.

As for the second part, by Inclusion-Exclusion, $\mathrm{Pr}[\text{choosing the captain} \cup \text{choosing the vice}]$ is equal to $$\overbrace{\frac{\binom{6}{2}}{\binom{7}{3}}}^{\mathrm{Pr}[\text{choosing the captain}]}+\overbrace{\frac{\binom{6}{2}}{\binom{7}{3}}}^{\mathrm{Pr}[\text{choosing the vice}]}-\overbrace{\frac{\binom{5}{1}}{\binom{7}{3}}}^{\mathrm{Pr}[\text{choosing the captain} \cap \text{choosing the vice}]}=\frac{5}{7}.$$

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  • $\begingroup$ Oh you're right, I calculated wrong! But what is the probability that person A, B or C are selected? Isn't this greater than 1? $\endgroup$ – user1776100 Sep 21 '13 at 23:03
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    $\begingroup$ In that case, the numerator is $\binom{3}{1} \times \binom{6}{2}-\binom{3}{2} \times \binom{5}{1}+\binom{3}{3} \times \binom{4}{0}=31$; so the probability is $31/35$. The probability shouldn't ever go greater than $1$ (if it did, it would just imply we've miscounted the number of combinations with a certain property). $\endgroup$ – Rebecca J. Stones Sep 21 '13 at 23:08

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