# Thread: Probability Real Life Question

1. ## Probability Real Life Question

So I am in my last year of pharmacy school and upon graduation I will be working for the Federal Bureau of Prisons....as a prison pharmacist. Long story short, there are 98 federal prisons, 22 of which I wouldn't mind being placed at (for various reasons), but each year only about 10 prisons have vacancies. So therefore, I would have to hope that one of my 22 preferred prisons is in that list of 10 job openings.

That's the story, here's the simplified question: What is the probability that if 10 numbers are chosen between 1-98, that at least one of them is 22 or below?

I'm not in any type of math class, just looking for help. I calculated it to be upwards of 90% but I'm interested to see what other people come up with. Let me know if anyone needs further clarification.

Thanks!

2. Originally Posted by byedabeast
So I am in my last year of pharmacy school and upon graduation I will be working for the Federal Bureau of Prisons....as a prison pharmacist. Long story short, there are 98 federal prisons, 22 of which I wouldn't mind being placed at (for various reasons), but each year only about 10 prisons have vacancies. So therefore, I would have to hope that one of my 22 preferred prisons is in that list of 10 job openings.

That's the story, here's the simplified question: What is the probability that if 10 numbers are chosen between 1-98, that at least one of them is 22 or below?

I'm not in any type of math class, just looking for help. I calculated it to be upwards of 90% but I'm interested to see what other people come up with. Let me know if anyone needs further clarification.

Thanks!
You should look at the probabilty of "bad luck", i. e., the probability that all the $10$ numbers are above $22$. If that probability is $p$, the probability you want is the complement, $1-p$.

For the first number, there are $98$ possibilities, and $98 - 22 = 76$ are bad; this corresponds to a probability of $\frac{76}{98}$.

Once the first number has been chosen, there are $97$ numbers left, and $97-22=75$ of them are bad; this corresponds to a probability of $\frac{75}{97}$. You can contine in this way until $10$ numbers have been picked.

The probability that all the $10$ numbers are above $22$ is the product of all these probabilities:

$\displaystyle\qquad p = \frac{76}{98}\cdots\frac{67}{89}\simeq 0.068$

and the probability that at least one of the numbers is less than or equal to $22$ is $1-p\simeq 0.932$.

Good luck

3. Originally Posted by byedabeast
So I am in my last year of pharmacy school and upon graduation I will be working for the Federal Bureau of Prisons....as a prison pharmacist. Long story short, there are 98 federal prisons, 22 of which I wouldn't mind being placed at (for various reasons), but each year only about 10 prisons have vacancies. So therefore, I would have to hope that one of my 22 preferred prisons is in that list of 10 job openings.

That's the story, here's the simplified question: What is the probability that if 10 numbers are chosen between 1-98, that at least one of them is 22 or below?
Is it reasonable to assume that every value (between one and ninety-eight) is equally likely? I mean, there's probably a reason why some locations are preferred over others, and this may cause a much lower turnover rate at those locations. It may be necessary to find some way to account for this variability.

4. Originally Posted by stapel
Is it reasonable to assume that every value (between one and ninety-eight) is equally likely? I mean, there's probably a reason why some locations are preferred over others, and this may cause a much lower turnover rate at those locations. It may be necessary to find some way to account for this variability.
Well, that's a good point.

To do that, we would need an estimate of the probability of vacancy in each prison. As such information was not included in the post, I used the "principle of indifference" by default.

Ideally, we should assign to each prison a weight proportional to the likelihood of a vacancy in that prison. Even if we had these weights, it would be hard to compute an exact solution, since we do not have independent events. We would maybe have to use a numerical Monte-Carlo simulation, since there are about $1.4\cdot 10^{13}$ possible outcomes.

We could try to find an approximate probability, by assigning an average weight $a$ to the "good" prisons, and an average weight $b$ to the "bad" prisons (only the ratio $(a:b)$ matters). These weights could be based on turnover statistics, for example.

With that approach, the probability of bad luck would be:

$\displaystyle \qquad p=\prod_{k=0}^9\frac{(76-k)b}{(76-k)b+22a}$

For example, with $(a:b)=(1:2)$, we get $p\simeq 0.239$. However, I don't know how good an approximation that would be. It could make sense if:

• The weights do no vary too much in each class.
• The averages are substantially different.

These conditions are typically what you would test using a correlation test like $\chi^2$. Note that this model assumes (pending confirmation) that the desirability of the prisons (from the pharmacist's point of view, of course) is related to the turnover rate. We do not know the OP's criteria; there may be other things involved, like distance from home, for example.