There are $N = k*k$ items.
An item is special with a probability of $p$ (uniform distribution).
Dividing the $N$ items into $k$ groups, each group with $k$ items.
Called a special group is a group with at least one special item.
What is the expected number of the special groups?
For instance, consider 10.000 items with the special rate of $1\%$. Dividing the $10.000$ items to $100$ groups, each group has $100$ items.
I am confused by two reasoning:
The probability of special group is $1\%$.
The probability of special item in each group is $1\%$, so we expect each group has 1 special item, so all groups are special groups, so the expected number of special groups is $100$.
Both solutions might be wrong.
What you are saying is that the average of the number of special items in a group is 1. But it does not mean that a group has certainly a special item.
The solution is as follows: The probability that a group has no special item is $q:=(1-p)^k$ (see the binomial distribution). So the number of special groups is also a binomial random variable with parameters $(k,1-q)$. So the expected number of special groups is $k(1-q)=k(1-(1-p)^k)$. In your example, it is approximately $100\times (1-\frac 1 e)$.