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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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