Everyone has 2 copies of the A/B/O gene. One gene is inherited from the mother and the other from the father. Everyone has a 1/2 probability to pass a randomly selected gene copy to their children.
Choosing to focus on A and O, people with AA or AO genes have blood type A; those with OO have type O blood. Suppose Eve and both her parents have blood type A, but her sister Eva has blood type O. Eve marries Adam, who has blood type O.
(a) If Eve and Adam’s first child had blood type A, what is the probability that Eve carries an O gene?
(b) If Eve and Adam’s first child had type A blood, what is the probability that their second child will as well?
So far I've done:
Let A = Eve carries an O gene, and C = first child has type A blood
P(A) = 6/12 = 1/2 because Eve's parents have to be either AO/AO, AO/AA, AA/AO (they cannot be AA/AA because Eve's sister has blood type O)
P(C) = 6/8 = 3/4 because parents have to be AA/OO or AO/OO.
I believe part a) is asking for P(A|C) = $\frac{P(C|A)P(A)}{P(C)}$
However, I'm not sure how to calculate P(C|A) or the probability that the first child has type A blood given Eve carries an O gene.
Best Answer
For (a), let the events be marked as:
$$A \sim \text{Eve having the O genes, i.e, (A, O)}$$ $$B \sim \text{1º child being type A, i.e, (A, O)}$$
By Bayesian: $$P(A|B) = \frac{P(B)P(B|A)}{P(A)}$$
We first calculate each term:
For the event B, we must think that, Eve will either pass gene A or pass the other gene. And for the case of sucess, the other gene must also be A. ALSO, both parents must be (A,O), otherwise Eva would never be (O,O) So...
$$P(B) = \frac{1}{2} + (\frac{1}{2}*\frac{1}{2}) = \frac{3}{4}$$
the event B|A is trivial, because is this scenario Eve is (A, O):
$$P(B|A) = \frac{1}{2}$$
also A:
$$P(A) = \frac{1}{2}$$
So...
$$P(A|B) = \frac{\frac{3}{4}*\frac{1}{2}}{\frac{1}{2}} = \frac{3}{4}$$
For the (b) I think that now you can do it by yourself.