Poisson Distribution

p(2) = 0.1839 ~ 2. There is about 1 chance in 5 that a second ambulance will be needed.

Explanation

The basic rate is r = 1 (in hour units), and we need to find the chance of getting 2 per hour. Since the rate is simple, we can find it on the Poisson Table under r = 1. We then read down to the p(2) row, and find that the corresponding probability is 0.1839. That is all there is to it.

Comment

The problem has been carefully worded to make it reasonably valid. Accident rates for the Stockholm Freeway over one year would include vastly different ice conditions, and this factor would make predictions for any single day questionable. Even in equable California, days differ from each other as to traffic: Monday is not invariably a weekday, and thus Tuesday is sometimes the first day of the work week, so morning rush hour on Wednesday is more likely to have a stable accident rate. It is vital to the validity of the Poisson distribution that its sole determinant, its rate r, is constant for the time being considered.

As for the hospital dispatcher, Poisson or no Poisson, she will want to be especially well provided with serviced and ready ambulances on (forbid the thought) a rainy day.