# Conditional Probability

I was able to do the first part of this problem but unsure how to approach this:

Mendel, revisited: Mendel’s peas had either purple or white flowers; flower color is due to a single gene, for which the purple allele (A) is dominant to the white allele (a). We cross two pure-breeding lines (one purple and one white) to produce the F1 hybrid. We self the F1 and choose an F2 seed at random. We grow and self the F2 and choose two F3 seeds at random. Consider the following events A1 = {F3 number 1 has purple flowers} and A2 = {F3 number 2 has purple flowers}.

(b) Are A1 and A2 conditionally independent, given F2? In other words, is Pr(A1 and A2| F2) = Pr(A1|F2) × Pr(A2| F2)?

Say for instance if F2 where heterozygote (Aa); ......AA means homozygote (purple) and aa (white)

I know:

P(A1 and A2) = P(A1 and A2 | Aa) P(Aa) + P(A1 and A2 | AA) P(AA) + P(A1 and A2 | aa) P(aa)
=(3/4)^2*(1/2)+1*(1/4)+0*(1/2)= 0.53
P(A1) = P(A2) = P(A1 | Aa) P(Aa) + P(A1 | AA) P(AA) + P(A1 | aa) P(aa)
= (3/4)*(1/2) + 1*(1/4) + 0*(1/4) = 0.625

chiro
Hey dspampi and welcome to the forums.

Have you tried instead of first calculating P([A1 AND A2]|F2) and then calculating separately P(A1|F2) and P(A2|F2)?

So for P([A1 AND A2]|F2) = P(A1 AND A2 AND F2)/P(F2) and P(A1|F2) = P(A1 AND F2)/P(F2) and P(A2|F2) = P(A2 AND F2)/P(F2).

I don't know anything about your biological probabilities, but you may have to do it by exhaustion. In other words do it for all possible values of F2 if you have some non-general distribution (You basically evaluate all of the probabilities for all possibilities of the random variable F2).

It would help if you posted what your probabilities are for each event for your random variables.

It looks like you have done the exhaustion for P(A1 and A2) but you haven't included the exhaustion for the F2 random variable. Doing this and combining it with your other results should give you the answer you need.