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QUESTION
A certain disease is only suffered by men and can only be
transmitted by direct inheritance from one's mother if she is a
carrier. If a woman is a carrier the probability that she
transmits the disease to any of her children ( the boys being
sufferers and the girls carrier), is 0.5, independently of all
other children. A woman knows that her brother has the disease.
If she already has two normal sons, what is the probability
that her next child will be unaffected?
ANSWER
Brother has disease $\Rightarrow$ Woman's mother carrier
$\Rightarrow$ P(woman carrier)=P(woman not
carrier)=$\frac{1}{2}$\\
P(two normal son's |carrier)=$\frac{1}{4}$, P(two normal son's
|not carrier)=1.\\
P(two normal sons)$=\frac{1}{2} \times \frac{1}{4} +\frac{1}{2}
\times 1=\frac{5}{8}.$\\
P(woman carrier| two normal)
$=\frac{\frac{1}{8}}{\frac{5}{8}}=\frac{1}{5}$, P(woman not
carrier|two normal sons)$=\frac{4}{5}$\\
P(next child normal|carrier)$=\frac{1}{2}$, P(next child
normal|not carrier)=1. Hence P(next child normal)$=\frac{1}{2}
\times \frac{1}{5} +1 \times \frac{4}{5} =\frac{9}{10}$
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