Consider a family with two male siblings, where both siblings are affected with

Question

Consider a family with two male siblings, where both siblings are affected with a genetically inherited disease. Suppose that – although the genetic history of the family is unknown – only dominant, recessive, or sex-linked modes of inheritance are possible, and each is equally likely (Probability of each = 1/3). After doing some research, you find the probability of this disease given a dominant, recessive, or sex-linked mode of inheritance are 0.25, 0.0625, 0.25, respectively.

Compute the likelihood of each mode of inheritance for a child who has the disease.

For the problem below, I am assuming that there is a Pa1=Pa2=Pa3 =0.33, this from the modes of inheritance. And then there is a P(b | a1) = 0.25, P(b | a2)=0.0625, and P(b|a3) = 0.25 on the probability of the kid having the disease based on their mode of inheritance.

How would the likelihood of each mode of inheritance be computed given that information?

Explanation / Answer

We can simply use Baye's theorem here.

P(A1|B) = P(B|A1)/(P(B|A1) + P(B|A2) + P(B|A3)) => Likelihood of dominant inheritance

P(A2|B) = P(B|A2)/(P(B|A1) + P(B|A2) + P(B|A3)) => Likelihood of recessive inheritance

P(A3|B) = P(B|A3)/(P(B|A1) + P(B|A2) + P(B|A3)) => Likelihood of sex-linked inheritance

substituting values, (using P(B|A1) + P(B|A2) + P(B|A3) = 0.25+0.0625+0.25 = 0.5625)

P(A1|B) = 0.25/0.5625 = 0.4445

P(A2|B) = 0.0625/0.5625 = 0.1111

P(A3|B) = 0.25/0.5625 = 0.4445

Please upvote the answer. Thank you!

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