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)) ?

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.

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 $\dfrac{P}{\binom{14}{4}}$.