 ### Description

If a woman is the daughter of an obligate carrier for an X-linked recessive disorder, then the more unaffected sons she has, the lower the probability that she is a carrier for the condition.

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.