Genetic diseases probability help

[mirandasatterley]

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?

 

[D H]

First a few comments:

the father cannot transmit the disease to the child, even if he is a carrier

This was true in the past because males used to die before they could father a child. The daughters of male hemophiliacs are carriers, and may be female hemophiliacs if the mother is a carrier. The status of male offspring of male hemophiliacs depends solely on their mother.

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).

You have the probability correct but not the reason. A female carrier has one normal and one defective X chromosome. The probability that she will pass the defective chromosome to anyone of her offspring is 50%.

What is the probability that a women is a carrier, given that she has one normal son?

Better stated as what is the probability that the daugher of a known carrier is also a carrier, given that ...

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

This is a conditional probability. The probability that the woman passes the disease to her son is zero if she is not a carrier. Thus you should use Pr(S|C) rather than Pr(S).

So, the probability that the women is a carrier, given that she has one normal son(complement of S) would be:

Do you know Bayes' Law?
I will denote the birth of a normal son as ~S. Therefore you want to calculate the prior probability Pr(C|~S). Apply Bayes' Law.

 

[mirandasatterley]

So, since the woman cannot pass on the disease if she doesn't have it:
Pr(S/C^c) = 0, and Pr(~S/C^c) = 0

If the woman has the disease,then the prob. that she has a son with the disease is:
Pr(S/C) = 0.50
And the prob. that she has a son without the disease is:
Pr(~S/C) = 1- 0.50 = 0.50

And if the probability that she passes it on is: Pr(C) = 0.50, then Pr(C^c) = 1- 0.5 = 0.50


According to Bayes' Law:

Pr(C/~S) = [Pr(Cn~S)] / [Pr(~S)]
= [Pr(~S/C) P(C)] / [Pr(~S/C) Pr(C) + Pr(~S/C^c) Pr(C^c)]
= [0.50 (0.50)] / [0.50(0.50) + 0(0.50)]
= 0.25/0.25
= 1
This answer doesn't make sense to me, but I can't see where I went wrong.

 

[D H]

This answer doesn't make sense to me, but I can't see where I went wrong.

Your mistake is in bold.

According to Bayes' Law:

Pr(C/~S) = [Pr(Cn~S)] / [Pr(~S)]
= [Pr(~S/C) P(C)] / [Pr(~S/C) Pr(C) + Pr(~S/C^c) Pr(C^c)]
= [0.50 (0.50)] / [0.50(0.50) + 0(0.50)]
= 0.25/0.25
= 1

The probability that a non-carrier mother will have a non-diseased son is not zero.

 

[mirandasatterley]

Okay - it's 1 right?
Thank you for all of your help!