Assumptions:
(1) The obligate carrier parent for the woman has only 1 copy of the defective X-linked gene.
(2) The risk of inheritance is 0.5.
The prior risk if there are no affected sons is 0.5 for being a carrier and for being a non-carrier.
conditional probability for being a carrier if she has normal sons =
= (0.5)^(number of normal sons)
conditional probability for not being a carrier if she has normal sons =
= 1
posterior probability of being a carrier =
= (0.5 * ((0.5)^(number of normal sons))) / ((0.5 * ((0.5)^(number of normal sons)))+ 0.5)
posterior probability of being a noncarrier =
= (0.5 * 1) / ((0.5 * ((0.5)^(number of normal sons)))+ 0.5) =
= 1 - (posterior risk of being a carrier)
If a son is affected, then she is likely a carrier, but the possibility of a new mutation needs to be considered.
