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mirandasatterley
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Homework Statement
Some genetic diseases (e.g. hemophilia, muscular dystrophy) can be transmitted to the child only if the mother is a carrier; the father cannot transmit the disease to the child, even if he is a carrier. This type of disease affects only males, carrier females rarely
exhibit any symptoms. In some situations, it can be determined that the probability that
a woman is a carrier for the disease is 1/2 (for instance if the woman is the daughter of a
known carrier of the disease). The probability that she will pass this disease to a son is also 1/2 (since the disease cannot be passed from father to son). The birth of a normal son adds evidence, but certainly not conclusive evidence, that the mother is not carrier of the disease gene. What is the probability that a women is a carrier, given that she has one normal son?
Homework Equations
Pr(A/B) = Pr(AnB)
Pr(B)
The Attempt at a Solution
Let C - That a woman is a carrier for the disease.
Therefore, Pr(C) = 0.5
Let S - The probability that a woman passes the disease to her son
Therefore, Pr(S) = 0.5
So, the probability that the women is a carrier, given that she has one normal son(complement of S) would be:
Pr(C/S^c) = Pr(CnS^c) ,
Pr(S^c)
Where Pr(S^c) = 1- Pr(S) = 1-0.5 = 0.5
How is Pr(C n S^c) calculated?