2.6.1 problem 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)

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

(b)

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