Suppose that there are 7 soccer players and 6 football players on a bus. Find the probability that a group of 5 players selected at random will consist of 3 soccer players and 2 football players.
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From soccer players $\displaystyle \binom{3}{7}$ From football players $\displaystyle \binom{2}{6}$ In total $\displaystyle \binom{5}{13}$
Hi,Thank you for the help! the answer is the back of the book says it is .408. Would you mind explaining it please?
Originally Posted by kangta27 Hi,Thank you for the help! the answer is the back of the book says it is .408. Would you mind explaining it please? $\displaystyle \frac{\dbinom{7}{3}*\dbinom{6}{2}}{\dbinom{13}{5}} \approx 0.408$
Originally Posted by pickslides From soccer players $\displaystyle \binom{3}{7}$ From football players $\displaystyle \binom{2}{6}$ In total $\displaystyle \binom{5}{13}$ Hmmm ... each of those equals zero.
Originally Posted by alexmahone Hmmm ... each of those equals zero. As you can see, I'm having one of my upside down days!
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