Chances of being ill on Fridays Background:
My son has been ill a few times this academic year and has missed some school.  The authorities get involved when attendence is low, and they're basically suggesting he's malingering, which he's not.
So one of the things they're implying is that he's been off a suspicious number of Fridays. I'd like to produce some stats to show it's just a statistical fluke
The problem statement is something like:
Given a student could have attended school on N days,
but was absent through illness on M of those days, 
what's the probability that they were absent for more than P Fridays.
As a follow up, what's the change that they were absent for more than P days for any one of the days of the week, not necessarily Friday.
Pointers to how to work this out for myself also welcomed.
 A: You can evaluate this by running a chi squared or Fisher's exact test on a 2x2 contingency table. Make the columns of the table "Sick" and "Not sick" and the rows "Friday" and "Not Friday". Now fill in the table by counting the number of instances of each case - Sick+Friday, Sick+Not Friday, Not sick+Friday, Not sick+Not Friday. Run a chi squared test on this table, and it will tell you if it's significantly different from the expected distribution (which would be that 20% of sicks days fall on Friday).
A: The crudest way I can think of it is this. Certainly not realistic, but pure combinatorial reasoning is sometimes insightful and might even provide a reasonable approximation. Caveat: two consecutive days of sickness count as two separate illnesses.
You are picking $M$ days out of $N$, so you have $N\choose{M}$ possibilities. Out of these, you are interested in those in which $P$ are Fridays, which are $N/5$ (assuming for simplicity the $N$ days span full weeks only).
To count how many cases satisfy your conditions, you can pick the $P$ Fridays in advance, and then pick the rest sick days among the non-Friday days that are left to choose from.
So you have ${N/5}\choose{P}$ Friday configurations, and ${4N/5}\choose{M-P}$ remaining days to pick.
This works out to a probability of
$$
\frac{{{N/5}\choose{P}} {{4N/5}\choose{M-P}}}{N\choose{M}}
$$
This is the chance the poor thing will get sick on $P$ Fridays. You can now just sum over all values at least $P$ to get the desired probability.
$$
\sum_{i=P}^M\frac{{{N/5}\choose{i}} {{4N/5}\choose{M-i}}}{N\choose{M}}
$$
