Question

Consider the game of 'Let's Make a Deal' in which there are three doors (numbered $1,\, 2,\, 3$), one of which has a car behind it and two of which are empty (have "booby prizes"). You initially select Door $1$, then, before it is opened, Monty Hall tells you that Door $3$ is empty (has a booby prize). You are then given the option to switch your selection from Door $1$ to the unopened Door $2$. What is the probability that you will win the car if you switch your door selection to Door $2$?

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (C)

$S = \{1, \,2,\, 3\}$, where outcome $'i'$ means that the car is behind door $i$. Let $E =$ {Door $3$ is empty} $= \{1,\, 2\}$. The probability that you win by switching to Door $2$, given that he tells you Door $3$ is empty is
$P\left(\left\{2\right\}\,|\,\left\{1,\, 2\right\}\right) = \frac{P\left(\left\{2\right\}\,\cap\,\left\{1,\, 2\right\}\right)}{P \left(\left\{1,\, 2\right\}\right)}$
$= \frac{1/3}{2/3} = \frac{1}{2}$