Exercise 2.46

Let $S$ be the event of successfully getting the Car under the specified strategy. Let $C_i$ be the event that Door $i$ contains the Car. Let $A$ be the event that Monty reveals the Apple, and let $A_i$ be the event that Door $i$ contains the Apple.

(a)

$$\begin{array}{llll}\hfill P(S)& =P(S\cap C_1)+P(S\cap C_2)+P(S\cap C_3)+P(S\cap C_4)& \hfill & \\ \hfill & =P(C_1)P(S|C_1)+P(C_2)P(S|C_2)+P(C_3)P(S|C_3)+P(C_4)P(S|C_4)& \hfill & \\ \hfill & =\frac{1}{4}\ast 0+3\ast \frac{1}{4}\ast (p+q)\frac{1}{2}& \hfill & \\ \hfill & =3\ast \frac{1}{4}\ast \frac{1}{2}& \hfill & \\ \hfill & =\frac{3}{8}& \hfill & \end{array}$$

(b)

$$\begin{array}{llll}\hfill P(A)& =P(A\cap G_1)+P(A\cap A_1)+P(A\cap B_1)+P(A\cap C_1)& \hfill & \\ \hfill & =P(G_1)P(A|G_1)+P(A_1)P(A|A_1)+P(B_1)P(A|B_1)+P(C_1)P(A|C_1)& \hfill & \\ \hfill & =\frac{1}{4}p+0+\frac{1}{4}q+\frac{1}{4}q& \hfill & \\ \hfill & =\frac{1}{4}(1+q)& \hfill & \end{array}$$

(c)

$$P(S|A)=\frac{P(S\cap A)}{P(A)}=\frac{\frac{1}{4}\ast p\ast \frac{1}{2}+\frac{1}{4}\ast q\ast \frac{1}{2}}{\frac{1}{4}(1+q)}=\frac{\frac{1}{8}}{\frac{1}{4}(1+q)}$$