# The smaller the population, the higher likelihood of the medical staff shortage. Is it true?

The problem statement:

Suppose we have a city with 100 doctors and the population of 1 million. Probability that a person will need a doctor on a given day is $$1\%$$. What is the probability that there won’t be enough doctors on a given day?

• By "not enough doctors" I assume it is meant that on a given day, there will be more people ($$> 100$$) seeking medical advice than there are doctors that can provide it.

• We further assume that $$100$$ doctors are not included in the population.

Seems like the most reasonable option here is to use binomial random variable. Using wolfram alpha gives

$$\sum_{x=101}^{10^6}{10^6 \choose x}0.01^x0.99^{10^6-x} \approx 0$$

Just for the sake of it, I wanted to see what the result would be given that we have a population of $$10^4$$. To my surprise, the result (wolfram alpha) is

$$\sum_{x=101}^{10^4}{10^4 \choose x}0.01^x0.99^{10^4-x} \approx 0.47$$

Which implies that, the smaller population, the higher probability of medical staff shortage. But shouldn't the opposite be true? The larger the population, the more people might need to seek medical advice, hence the higher likelihood of not having enough doctors. What am I misunderstanding?

• This first calculation is wrong. No idea how Wolfram calculated the summation, but it's clear that on average, $0.01 \times 10^6 = 10,000$ people need a doctor in a day. Do you mean that probability is 0.01**%**? Apr 10, 2020 at 13:59
• @AdamO, I meant probability of $1\%$, or $\frac{1}{100}$ Apr 10, 2020 at 14:00
• Did you realize the 0 probabilty is wrong? It would be instructive to correct the probability calculation. You can use R and pbinom. Apr 10, 2020 at 14:01

I suspect you're running into computational issues with Wolfram Alpha trying to do an exact calculation with $$n = 10^6$$. I tried finding the complement of the probability you were looking for, figuring that would be simpler, and I got a notification saying the calculation timed out. Splitting up that calculation in to two parts (0-50 people need a doctor and 51-100 people need a doctor), I didn't get that error, and both probabilities were approximately 0, so combining them would also yield 0. This would then imply that the probability that more than 100 people need a doctor is essentially 1.
As AdamO mentioned in the comments above, we should expect an average of 10,000 people per day to need a doctor if $$p = 1%$$. If you compute $$P(x>100)$$ using a normal approximation to the binomial distribution, you should find this probability is basically 1.