# Probabibility problem in terms of binomial coeficciants

• Feb 25th 2014, 09:24 PM
crownvicman
Probabibility problem in terms of binomial coeficciants
From 6 positive integers and 8 negative integers, 4 are chosen at random without replacement. What is the probability that their product is positive? Express answer in terms of binomial coefficients

So I know the total is 14 integers, and for the product to be positive the selection has to be either all 4 are + , all 4 are - , or 2 are + & 2 are -

I'm confused as to how this would be expressed in terms of binomial coefficients. Would it be ((4 choose 6) + (8 choose 4)/(14 choose 4)) ?
• Feb 26th 2014, 12:13 AM
chiro
Re: Probabibility problem in terms of binomial coeficciants
Hey crownvicman.

You are better off selecting the independent events and adding up their probabilities if they are disjoint. This avoids trying to use formulae that may not work.
• Feb 26th 2014, 04:13 AM
Plato
Re: Probabibility problem in terms of binomial coeficciants
Quote:

Originally Posted by crownvicman
From 6 positive integers and 8 negative integers, 4 are chosen at random without replacement. What is the probability that their product is positive? Express answer in terms of binomial coefficients
So I know the total is 14 integers, and for the product to be positive the selection has to be either all 4 are + , all 4 are - , or 2 are + & 2 are -
I'm confused as to how this would be expressed in terms of binomial coefficients. Would it be ((4 choose 6) + (8 choose 4)/(14 choose 4)) ?

Let $P =\displaystyle \sum\limits_{k = 0}^2 {\binom{6}{2k}\binom{8}{4-2k}}$ then find $\dfrac{P}{\binom{14}{4}}$.