$24$ students arrive in random order to collect shirt, $3$sizes available $24$ students arrive in random order. $12$ are size $S$ and $12$ are size $M$. There are $24$ T-shirts provided: $10$ are size $S,$ $10$ are size $M,$ and $4$ are size $L.$ A student takes their correct size if it is available, or failing that the next size up. They go without if that is also not available. What is the probability that all students get T-shirts?
[Someone else posted this problem and I was busy working on the solution when they deleted it. It is quite an interesting problem, so I decided to repost it.]
 A: Everyone gets a shirt unless the medium shirts are gone before all of the small people have had a chance to collect a shirt.
Worst case scenario.  The first $10$ people to arrive are mediums, $4$ people go home without shirts.
We only have to look at the shirt sizes of the last $4$ people who show up.
By the time the $20^{th}$ person arrives all of the mediums will be gone.
If any small people are in the last $4$ to arrive one of them doesn't collect a shirt.
The chance that the last $4$ are all mediums is $\frac{12\choose4}{24\choose4} = 0.0466$
A: All "mediums" are sure to get T-shirts.
All "smalls" will get suitable T-shirts only if any $4$ of the "mediums" are at the end,
which has the same probability as any $4$ of the mediums being at the start,
since all permutations are equi-probable.
Thus $Pr = \frac{12\cdot11\cdot10\cdot9}{24\cdot23\cdot22\cdot21}$
A: Label the size $S$ students in order of arrival as $S_1,\dots,S_{12}$ and similarly the size $M$ students in order of arrival as $M_1,\dots,M_{12}$. The $M$ students are bound to get a shirt, but $S_{11},S_{12}$ may find that all available $S$ and $M$ shirts have been taken. At least one of them goes without a shirt iff $M_9$ arrives before $S_{12}$.
If the last four students to arrive are all size $M,$ then all earlier students must have got a shirt, because the other $8$ size $M$ would only take $8$ of the size $M$ shirts leaving $2$ for the $12$ size $S$ students. So the last four students all take size $L$ and every student gets a shirt.
If $S_{12}$ arrives in the last four, then $M_9$ must already have arrived, so $S_{12}$ fails to get a shirt.
There are ${24\choose12}$ possible arrival patterns, all equally likely. The number with the last four arrivals being $M$ is ${20\choose12}$. So the required probability is $\frac{15}{322}$.
