Your workplace has a phone room for employees to quietly make personal calls. The room has no windows, just a sign that can be switched from “Vacant” to “Occupied”. However, employees differ in how consistently they use the sign:

- 1/2 of them always switch to “Occupied” when they enter and “Vacant” when they exit.
- 1/4 of them ignore the sign altogether – the sign will always read the same before, during, and after their visit.
- 1/4 of them always switching to “Occupied” when they enter, but always forget to switch back to “Vacant” when they exit.

If the room is actually occupied exactly 1/2 of the time, what is the probability the room is actually vacant when the sign reads “Vacant”?

#### Solution

4/5 or 80%

This is just some good old-fashioned conditional probability: P(vacant | sign reads “Vacant”) = P(vacant AND sign reads “Vacant”) / P(sign reads “Vacant”).

If the sign reads “Vacant”, one of these must be true:

- The room is vacant AND the last employee who doesn’t ignore the sign was an employee that switches on both enter and exit (otherwise the sign would read “Occupied”).
- Room is vacant = 1/2
- Last employee who doesn’t ignore the sign was an employee that switches on both enter and exit = (1/2) / (1/2 + 1/4) = 2/3
- Both are true = 1/2 x 2/3 = 1/3

- The room is occupied by an employee that ignores the sign AND the last employee who doesn’t ignore the sign was an employee that switches on both enter and exit.
- Room is occupied by an employee that ignores the sign = 1/2 x 1/4 = 1/8
- Last employee who doesn’t ignore the sign was an employee that switches on both enter and exit = (1/2) / (1/2 + 1/4) = 2/3
- Both are true = 1/8 x 2/3 = 1/12

So the probability the room is actually “Vacant” = (1/3) / (1/3 + 1/12) = 4/5.