Confusing probability question I have got a task, which seems a quite confusing for me. It is simple: In a market, they sell eggs in egg holders, they store $10$ of them in each. There is $60$% chance, that all of the eggs are ok, $30$% chance, that exactly $1$ of them is broken, and $10$% chance, that exactly $2$ of them are broken(it is random, which one is broken).
We buy an egg holder, and after we grab our first egg, we are sad, because it is broken. What is the probability, that there is one more broken egg in our holder?
The "logical" way would be: $30$% of them have $1$ broken egg, $10$% of them have $2$, so, to have $2$ broken, the chance must be $\frac14$. But I am not really sure if that is the correct approach, since the broken egg can be anywhere, getting a broken one for first may be not that easy, or is that independent?(Maybe, I could use Bayes Theorem somehow)?
Any help appreciated.

Not really an answer, but more a visualisation of the excellent answer of Lulu.
$\begin{array}{cccccccccc}
A & A & A & A & A & A & B & B & B & C\\
0 & 0 & 0 & 0 & 0 & 0 & \otimes & \otimes & \otimes & \otimes\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \otimes\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}$
The columns correspond with boxes. You picked a broken egg. The $5$ broken eggs have equal chances to be the elected one. Exactly $2$ of the $5$ are located in a box that contains another broken egg.
