Exercise 2.60

Let $D$ be the event that a person has the disease. Let $T$ be the event that a person tests positive for the disease.

(a)

$$\begin{array}{llll}\hfill P(D|T)& =\frac{P(D)P(T|D)}{P(T)}& \hfill & \\ \hfill & =\frac{p(P(A|D)P(T|D,A)+P(B|D)P(T|D,B))}{P(A)P(T|A)+P(B)P(T|B)}& \hfill & \\ \hfill & =\frac{p\left(\frac{1}{2}{a}_1+\frac{1}{2}{b}_1\right)}{\frac{1}{2}\left(P(D|A)P(T|D,A)+P(D^c|A)P(T|D^c,A)\right)+\frac{1}{2}\left(P(D|B)P(T|D,B)+P(D^c|B)P(T|D^c,B)\right)}& \hfill & \\ \hfill & =\frac{\frac{1}{2}p(a_1+b_1)}{\frac{1}{2}(p{a}_1+(1-p)(1-a_2))+\frac{1}{2}(p{b}_1+(1-p)(1-b_2))}& \hfill & \end{array}$$

(b)

$$\begin{array}{llll}\hfill P(A|T)& =\frac{P(A)P(T|A)}{P(T)}& \hfill & \\ \hfill & =\frac{\frac{1}{2}(p{a}_1+(1-p)(1-a_2))}{P(A)P(T|A)+P(B)P(T|B)}& \hfill & \\ \hfill & =\frac{\frac{1}{2}(p{a}_1+(1-p)(1-a_2))}{\frac{1}{2}(p{a}_1+(1-p)(1-a_2))+\frac{1}{2}(p{b}_1+(1-p)(1-b_2))}& \hfill & \end{array}$$