I am confused on the below problem. I think conditional probability factors in but I'm not exactly sure how to solve the condition as, using this $$ P(B\mid A) = \frac{P(B)\cdot P(A\mid B)}{P(A)}$$ does not seem to help as it is just as hard conditioned the opposite way around is just as difficult. Any help is appreciated. Thank you. Also, B does not factor into the problems for simplicity. We're just supposed to think of A and O alleles.

As you may remember from basic biology, the human A/B/O blood type system is controlled by one gene for which 3 variants (“alleles”) are common in the human population – unsurprisingly called A, B, and O. As with most genes, everyone has 2 copies of this gene, one inherited from the biological mother and the other from the biological father, and everyone passes a randomly selected copy to each of their biological children (probability 1/2 for each copy, independently for each child). Focusing only on A and O, people with AA or AO gene pairs have type A blood; those with OO have type O blood. (A is “dominant”, O is “recessive”.) Suppose Apple and both of her biological parents have type A blood, but her biological sister Olive has type O.

Apple and a man with type O blood have two children. If their first child has type A blood, what is the probability that their second child will as well?

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    $\begingroup$ Hint: You know from the sister that Apple's parents each have AO $\endgroup$ – Henry Apr 28 '20 at 22:41
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    $\begingroup$ To start, have you worked out the prior probability Apple has genes AA or AO before considering the information about her first child? The conditional probability that Apple's child has blood type A or O given that Apple has genes AA and given that Apple has genes AO? $\endgroup$ – aschepler Apr 28 '20 at 22:47

You don't have to do this all in one fell swoop. Take little bites.

Only A and O alleles figure in. Since Alice's parents are each type A, each has an A allele. Since Olive is type O, each parent donated an O allele. Therefore, each parent is AO. Alice's mate is OO, so there are no B alleles involved.

What are the probabilities for Alice prior to her first blood test?

  • OO: $\frac14$
  • AO: $\frac12$
  • AA: $\frac14$

When Alice turn outs to have type A blood, the first possibility is rule out, and the probabilities become AO:$\frac23$ and AA:$\frac13$. Now let $A$ be the event that Alice's genotype is AA, and let $C$ be the event that her first child has type A blood.

The posterior probability that Alice is type AA will be $\Pr(A|C)$ since event $C$ comes to pass. By Bayes' rule, $$\Pr(A|C) = \frac{\Pr(A)\Pr(C|A)}{\Pr(C)}$$ Now, if Alice is genotype AA, then her child will certainly be type A, so $\Pr(C|A)=1$, and we know that $\Pr(A)=\frac13$ so $$\Pr(A|C)=\frac1{3\Pr(C)}$$

To figure out $\Pr(C)$, just use the law of total probability $$\Pr(C)=\Pr(A)\Pr(C|A)+\Pr(A^c)\Pr(C|A^c)$$ (By $A^c$ I mean the complement of $A$, that is, Alice's genotype is AO.)

Once you have updated the probability that Alice's genotype is AA, you can use the law of total probability another time to figure out the probability that the second child has type A blood.


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