In a game show, a contestant is shown three closed doors: Door A, Door B and Door C. Behind one…

Let $A, B, C$ denote the events that the laptop is behind Door A, Door B, and Door C respectively. $P(A) = P(B) = P(C) = \frac{1}{3}$. Let $D_B$ be the event that the host opens Door B. - If the laptop is behind A, the host can open B or C with equal probability: $P(D_B|A) = \frac{1}{2}$. - If the laptop is behind B, the host never opens B: $P(D_B|B) = 0$. - If the laptop is behind C, the host cannot open A (contestant's pick) and cannot open C (contains laptop), so must open B: $P(D_B|C) = 1$. By Bayes' theorem: $P(A|D_B) = \frac{P(A) P(D_B|A)}{P(A)P(D_B|A) + P(B)P(D_B|B) + P(C)P(D_B|C)}$ $= \frac{\frac{1}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 1} = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{3}} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}$.

By Bayes' theorem: $P(C|D_B) = \frac{P(C) P(D_B|C)}{P(A)P(D_B|A) + P(B)P(D_B|B) + P(C)P(D_B|C)}$ $= \frac{\frac{1}{3} \cdot 1}{\frac{1}{6} + 0 + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$.

The contestant should change their selection to Door C because the probability of winning the laptop by switching to Door C is $\frac{2}{3}$, which is twice the probability of winning by staying with Door A ($\frac{1}{3}$).