Since hypergeometric distribution is the without replacement version of the Binomial distribution why can't we replace the combinations in the Hypergeometric PMF with combinations with replacement and expect the same result with the standard Binomial PMF ?
To elaborate this is the regular combination formula counting unordered without replacement :
$$C(n,r)= \frac{n!}{r!(n−r)!}$$
And this is the combination with replacement formula counting unordered with replacement:
$$C^R(n,r)= \frac{(n+r−1)!}{r!(n−1)!}$$
and this is the hypergeometric distribution PMF:
For $X\sim\operatorname{HGeom}(w,b,n)$
$$P(X=k)= \frac{C(w,k)C(b,n-k)}{C(w+b,n)}$$
My question is, since binomial and hypergeometric are both fixed number of trials where Binomial is counting with replacement and Hypergeometric is counting without replacement, why doesn't altering the Hypergeometric formula to be
$$P(X=k)= \frac{C^R(w,k)C^R(b,n-k)}{C^R(w+b,n)}$$
give us the PMF of the Binomial Distribution ? What am I missing ?
Best Answer
If you have for example a bag with $6$ red balls and $4$ blues balls and you want the probability that drawing $3$ balls gives you $2$ red and $1$ blue, then
which are very different calculations and there is no real correspondence between the two. The "with replacement" binomial version has a succession of products of small fractions which are either all the same or $1 $ minus the same term and this really cannot be consolidated into ratio of single binomial coefficients.