Exercise 2.44

Let $G_i$ be the event that the $i$-th door contains a goat, and let $D_i$ be the event that Monty opens door $i$. Let $S$ be the event that the contestant is successful under his strategy.

(a)

There are two scenarios which result in the contestant selecting door $3$ and Monty opening door $2$. Either the car is behind door $3$ and Monty randomly opens door $2$, or doors $3$ and $2$ contain goats, and Monty opens door $2$. Only the latter scenario results in a win for the contestant.

Thus,

$$P(S|D_2,G_2)=\frac{(p_1+p_2)\frac{p_1}{p_1+p_2}}{p_3\frac{1}{2}+(p_1+p_2)\frac{p_1}{p_1+p_2}}=\frac{p_1}{p_1+\frac{1}{2}{p}_3}.$$

(b)

We can slightly modify the scenario in part $a$ where doors $3$ and $2$ contain goats by multiplying the probability of the scenario by $\frac{1}{2}$ to accomodate the chance that Monty might open the door with the car behind it.

$$P(S|D_2,G_2)=\frac{\frac{1}{2}(p_1+p_2)\frac{p_1}{p_1+p_2}}{p_3\frac{1}{2}+\frac{1}{2}(p_1+p_2)\frac{p_1}{p_1+p_2}p}=\frac{\frac{1}{2}{p}_1}{\frac{1}{2}{p}_1+\frac{1}{2}{p}_3}=\frac{p_1}{p_1+p_3}.$$

(c)

$$P(S|D_2,G_2)=\frac{p_3}{p_3+\frac{1}{2}{p}_1}.$$

(d)

$$P(S|D_2,G_2)=\frac{p_3}{p_3+p_1}.$$