Occupied (Solution)

This is a fairly straightforward application of bayes theorom. To make things easier, let’s define a few symbols:

$o$ = bathroom is occupied
$v$ = bathroom is vacant
$\bar{o}$ = bathroom sign says occupied
$\bar{v}$ = bathroom sign says vacant

With this new terminology, let’s go through the question again:

Assume that 1/3 of bathroom users don’t notice the sign upon entering or exiting. Therefore, whatever the sign reads before their visit, it still reads the same thing during and after their visit. Another 1/3 of the users notice the sign upon entering and make sure that it says “Occupied” as they enter. However, they forget to slide it to “Vacant” when they exit. The remaining 1/3 of the users are very conscientious: They make sure the sign reads “Occupied” when they enter, and then they slide it to “Vacant” when they exit.

$$P[\bar{o}|o] = \frac{1}{3}P[\bar{o}|v] + \frac{2}{3}(1) \\ P[\bar{o}|v] = \frac{1}{3}P[\bar{o}|v] + \frac{1}{3}(1) + \frac{1}{3}(0) \\ \frac{2}{3}P[\bar{o}|v] = \frac{1}{3} \\ P[\bar{o}|v] = \frac{1}{2} \\ \therefore P[\bar{v}|v] = \frac{1}{2} \\ \therefore P[\bar{o}|o] = \frac{1}{3}\frac{1}{2} + \frac{2}{3}(1) = \frac{5}{6} \\$$

Finally, assume that the bathroom is occupied exactly half of the time, all day, every day.

$$P[o] = 1/2 \\ P[v] = 1/2 \\$$

Two questions about this workplace situation:

If you go to the bathroom and see that the sign on the door reads “Occupied,” what is the probability that the bathroom is actually occupied?

$$P[o|\bar{o}]?$$

Apply bayes theorem:

$$P[o|\bar{o}] = \frac{P[\bar{o}|o] P[o]}{P[\bar{o}]}$$

We know everything except $P[\bar{o}]$ but that’s easy to solve for:

$$P[\bar{o}] = P[\bar{o}|o]P[o] + P[\bar{o}|v]P[v] = \frac{5}{6}\frac{1}{2} + \frac{1}{2}\frac{1}{2} = \frac{2}{3}$$

So, plugging everything in:

$$P[o|\bar{o}] = \frac{P[\bar{o}|o] P[o]}{P[\bar{o}]} \\ P[o|\bar{o}] = \frac{\frac{5}{6} \frac{1}{2}}{\frac{2}{3}} = \frac{5}{8}\\$$

If the sign reads “Vacant,” what is the probability that the bathroom actually is vacant?

$$P[v|\bar{v}] = \frac{P[\bar{v}|v]P[v]}{P[\bar{v}]} \\ P[\bar{v}] = \frac{1}{2}\frac{1}{2} + \frac{1}{6}\frac{1}{2} = \frac{1}{3} \\ P[v|\bar{v}] = \frac{\frac{1}{2}\frac{1}{2}}{\frac{1}{3}} = \frac{3}{4} \\$$