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.

The answer isn't just $1/4$, because you're more likely to pick a broken egg first if the holder has two broken eggs.
Let $B$ be the event where the first egg we observe is broken, $H_1$ the event our holder has one broken egg and $H_2$ the event where our holder has two broken eggs.
Since there's one broken egg out of ten in the $H_1$ case, $P(B\ |\ H_1) = 1/10$. Likewise there are two broken eggs in the $H_2$ case, so $P(B\ |\ H_2) = 2/10$.
This means $P(B) = 1/10 \times 3/10\ +\ 2/10 \times 1/10 = 3/100 + 2/100 = 5/100$. You have a five percent chance of first examining a broken egg whenever you buy a holder of 10 eggs.
Now we're interested in $P(H_2\ |\ B)$ or the probability we have two broken eggs after we've observed the first one. By Bayes' theorem:
$P(H_2 | B) = \frac{P(B | H_2)P(H_2)}{P(B)} = \frac{(2/10)\ \times\ (1/10)}{5/100} = \frac{2/100}{5/100} = \frac{2}{5}$.
