You have 13 keys... and 5 key holes....
5 keys work in the 5 key holes...
In random chance what are the odds of getting randomly all 5 keys in the right key holes on the first atempt.
Bonus. getting 4right and 1 wrong on on the first attempt. And getting the one wrong one correct on a second atempt
Assuming that each of the 5 keys will unlock 1 specific lock, you can get the result in (at least) two different ways:
For each lock, choose a key at random: The chance that you choose the right key for the first lock is $\frac{1}{13}$. Choosing among the remaining keys, you will get the next key right with probability $\frac{1}{12}$, and so forth, giving you \begin{align*} \frac{1}{13}\cdot\frac{1}{12}\cdot\frac{1}{11}\cdot\frac{1}{10}\cdot\frac{1}{9}=\frac{1}{154440}. \end{align*}
Choosing 5 keys first, then place each in a lock: You can choose the 5 keys in $13\choose{5}$ ways. Afterwards, you can place each in a lock in $5!$ different ways. Only one of these are right, giving you a probability of \begin{align*} \frac{1}{{13\choose{5}} 5!}=\frac{1}{154440}. \end{align*}