so you have 1440 pee-minutes
available per day
60 minutes per hour, 8 hours = 480 minutes
* 3 stalls = 1440 total minutes
4:19 PM so if we assume that the average Jane pees three times a day, * 26 employees * 3 minutes per pee
you get 234 minutes of pee-time
and if you divide that by 1440 minutes of pee-time
4:20 PM you get a 16.25% percent likelihood that your restrooms will be at capacity at some point that day
4:21 PM so there you go:
me: I'm going to see if other people come up with the same answer :)
Rob: haha
well, keep in mind
I gave you a method
have them assume three pees per day, three minutes per pee
4:22 PM eight-hour workdays
I could be totally wrong on my math, too--I'm out of practice
me: uh
we work 9 hours.
Rob: okay
4:23 PM including lunch break?
me: no
10 hours including lunch break
Rob: most people leave during their lunch break?
hehe
we'd need the average work day
me: no
I'm talking about my company
Rob: call it 9.5
I know
4:24 PM I mean the average number of hours somebody is actually at the office
call it 9.5 to account for people leaving during the day, but also to account for people staying in and bringing their lunch?
me: I think your numbers are also messed up
k
4:25 PM Rob: assuming a longer day, it comes to about 13.7%
of course, that's assuming the pee times are equally distributed
4:26 PM and that we stay at an average of three pees per day

Well, I think this problem is more complicated than you are giving it credit for having.
This is far from the same as tossing three coins and seeing if they all come up heads. First there are an infinite amount of sample points during the day (time can be divided infinitely), and even if you were to divide them into countable increments, say minutes, then all sample points would not have the same probability of occurring. You are going to have a higher incidence of bathroom use after periods of liquid consumption (mid-morning, lunch, afternoon snack).
If you would further define the situation/question, I might be willing to spend the time to give you a real answer.

I agree that the problem isn't sufficiently defined.

Do engineers only use the unisex doored bathrooms, or do they choose randomly between all available bathrooms?

If someone is unable to use a bathroom at their chosen time because they're all full, do they wait in line until one is open and then go immediately?

To echo Rob and Tamara, do people pee at random times, or is there an after-lunch rush or other preferred time?

Assuming random pee times, I think the chance of all three women's stalls being full is less than the 16.25% that Rob predicts, for the reason Vivi mentions. Rob's number may be closer to the probability that there is at least one women's stall occupied.

I also agree that it's not well-defined. But to get a ballpark idea, throw out the males and the engineers and the unisex bathroom, and the lunch break, and assume the distribution is random, i.e. independent of lunch, etc., leaving this:

20 co-workers share a bathroom with three stalls. They each visit three times per day, with each visit lasting three minutes. What is the probability that, at any given moment during the eight-hour work day, all three stalls will be occupied?

So then you have 60 visits occurring over a range of 480 minutes. The probability of a visit starting during a given minute is 60/480, or 1/8.

Now, this will be approximate because, as Tamara alluded, we're discretizing what's really a continuous process, but to avoid Calculus...

In order for all three stalls to be occupied someone has to walk in first. At any given moment, this has probability 1/8. Then, two other people have to walk in during the three minutes that she's in there. Each of these has probability about 3/8 (in reality a bit lower since we should be using 59/360 and then 58/360).

These events are independent, so it's 1/8*3/8*3/8 = about 1.78%