multivariate hyper geometric Vs. Indep. drawing without replacement

I think my problem can be seen as a basic urn setup, with another layer tacked on after the fact.

I'm looking at drawing from an urn with 3 biological specimens. Each has genotype AA, Aa, or aa.

For this example let's assume there are two of each, but I'm supposed to code a where the quantities are our inputs.

We want to know, if we draw two items, and let them mate, what s the probability that the offspring will be aa. The offspring will recieve at random one letter from each parent.

Pr(offspring being aa if both parents are aa)= 1 Pr(offspring being aa if either parent is AA) = 0 Pr(offspring being aa if both parents are Aa) = 1/4 Pr(offspring being aa if one parent is aa and the other is Aa) = 1/2

So the total probability offspring being aa is a simple total law application. $$\sum_{i=pairing} pr(aa offspring|pairing)*pr(pairing)$$

The pr(aa_offspring| pairings) are listed above. I find the correct pr(pairing)'s, using a multivariate hypergeometric distribution. $$pr(aa paired Aa) = \frac{\binom{total aa}{1}\binom{total Aa}{1}}{\binom{urn_poulation}{2}}$$

In the numerator we don't consider AA's, because they never produce the desired offspring. But the denominator is 6! over 2!*4!, or 15 in our example because the AA's have to be accounted for in total population numbers.

I'm not getting correct answers using simple multiplication of probabilities of drawing certain types, without replacement, i.e.

Pr(aa paired Aa) = 2/6*2/5

By the time you fill out all the terms this in this (2 of each genotype) example, the $$pr(aa) = 13/30$$

I was just hoping someone could tell me if there is some underlying assumption or difference in these methods that makes the later unsuitable for the problem?Because they both seem valid to me. Or am I just screwing up the arithmetic somewhere and not seeing it?

Your value $\frac26\cdot\frac25$ doesn't account for that fact that you could draw aa and Aa in one of two orders, whereas the value in the earlier displayed equation does.
• @Carl: Yes, that's right. With that factor of $2$, you get $\frac4{15}$, which is also what you get from the displayed equation, $$\frac{\binom21\binom21}{\binom62}\;.$$ – joriki Jul 29 '16 at 20:12