# Combinatorial Prob.

• Jan 10th 2012, 07:38 PM
kangta27
Combinatorial Prob.
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.
• Jan 10th 2012, 07:54 PM
pickslides
Re: Combinatorial Prob.
From soccer players $\displaystyle \binom{3}{7}$

From football players $\displaystyle \binom{2}{6}$

In total $\displaystyle \binom{5}{13}$
• Jan 10th 2012, 08:08 PM
kangta27
Re: Combinatorial Prob.
Hi,Thank you for the help! the answer is the back of the book says it is .408. Would you mind explaining it please?
• Jan 10th 2012, 08:39 PM
alexmahone
Re: Combinatorial Prob.
Quote:

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?

$\displaystyle \frac{\dbinom{7}{3}*\dbinom{6}{2}}{\dbinom{13}{5}} \approx 0.408$
• Jan 10th 2012, 08:42 PM
alexmahone
Re: Combinatorial Prob.
Quote:

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. (Giggle)
• Jan 11th 2012, 01:15 AM
pickslides
Re: Combinatorial Prob.
Quote:

Originally Posted by alexmahone
Hmmm ... each of those equals zero. (Giggle)

As you can see, I'm having one of my upside down days!