2.3.3 problem 43

Let $C_i$ be the event that Door $i$ contains the car. Let $D_i$ be the event that Monty opens Door $i$. Let $O_i$ be the event that Door $i$ contains the computer, and let $G_i$ be the event that Door $i$ contains the goat.

(a)

$$\begin{array}{cccc}P\left(C_3\right|D_2,G_2)& =\frac{P\left(C_3\right)P(D_2,G_2|C_3)}{P(D_2,G_2)}& & \\ & =\frac{P\left(C_3\right)P(D_2,G_2|C_3)}{P\left(C_3\right)P(D_2,G_2|C_3)+P(C_3^c\left)P\right(D_2,G_2\left|C_3^c\right)}& & \\ & =\frac{\frac{1}{3}\ast \frac{1}{2}}{\frac{1}{3}\ast \frac{1}{2}+\frac{2}{3}P\left(G_2\right|C_3^c\left)P\right(D_2|G_2,C_3^c)}& & \\ & =\frac{\frac{1}{3}\ast \frac{1}{2}}{\frac{1}{3}\ast \frac{1}{2}+\frac{2}{3}\ast \frac{1}{4}}& & \\ & =\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{6}}& & \\ & =\frac{1}{2}.& & \end{array}$$

(b)

$$\begin{array}{cccc}P\left(C_3\right|D_2,O_2)& =\frac{P\left(C_3\right)P(D_2,O_2|C_3)}{P\left(C_3\right)P(D_2,O_2|C_3)+P(C_3^c\left)P\right(D_2,O_2\left|C_3^c\right)}& & \\ & =\frac{P\left(C_3\right)P(D_2,O_2|C_3)}{P\left(C_3\right)P(D_2,O_2|C_3)+P(C_3^c\left)P\right(D_2,O_2\left|C_3^c\right)}& & \\ & =\frac{\frac{1}{3}P\left(O_2\right|C_3\left)P\right(D_2|O_2,C_3)}{\frac{1}{3}P\left(O_2\right|C_3\left)P\right(D_2|O_2,C_3)+P\left(C{3}^c\right)P(D_2,O_2|C_3^c)}& & \\ & =\frac{\frac{1}{3}\ast \frac{1}{2}\ast p}{\frac{1}{3}\ast \frac{1}{2}\ast p+\frac{2}{3}\ast \frac{1}{4}\ast q}& & \\ & =\frac{\frac{1}{6}p}{\frac{1}{6}p+\frac{1}{6}q}& & \\ & =\frac{1}{2}p.& & \end{array}$$