The following starts off as a standard eye color problem... but I have a question at the end, so please be patient
The color of a person's eyes is determined by a single pair of genes. B = brown, b = blue
- if you have BB, Bb, bB, you get brown eyes
- if you have bb, you get blue eyes
and the chances of getting each individual gene are equal
the child gets 1 from the father, 1 from the mother to complete his pair
suppose Smith and both his parents have brown eyes, but Smith's sister has blue eyes.
a) What is the prob that Smith possesses one blue-eyed gene?
b) Suppose Smith's wife has blue eyes, what is the prob that their first child will have blue eyes?
c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?
The answer to a & b is 2/3 and 1/3. However, in c), it would seem that the events of giving birth to first and second children are independent, so the answer should be 1/3. But consider...
- suppose first child has blue eyes, that confirms Smith to have bB, which changes c) to 1/2
- how about, if the couple, say, has 10 kids, all of whom have brown eyes. how would that change (or not change) the underlying distribution of Smith's eye color genes?
ahh got it!
we know that after part b, Smith has the following distribution for genes
2/3: bB, 1/3: BB
his wife is 1/1: bb
P(smith_bB | child1_bB) = ( P(child1_bB | smith_bB) * P(smith_bB) ) / P(child1_bB) = (1/2 * 2/3) / (2/3) = 1/2
P(child2_bB | child1_bB)
= P(child2_bB & smith_bB | child1_bB) (since P(child2_bB & smith_bb | child1_bB) = 0)
= P(child2_bB | smith_bB & child1_bB) = P(child2_bB | smith_bB = 1/2) (and implicitly smith_BB = 1/2)
= 1/2*1/2 + 1/2 = 3/4