I have $k$ cows. I have $n$ fields. I release the cows and the cows randomly go to some field. Every field is equally likely to be picked. The number of fields may be less, equal, or greater than number of cows. Many cows can be in the same field.
What is expected number of cows that will visit each field.
How many fields are expected to be visited.
I have tried applying the following for question 1: Its $k/n$ I think.
Not sure at all about question 2.
We want the expected number of fields that have at least one cow in it.
To start, let's find the probability of a single fields having at least one cow in it, which is equivalent to $1$ minus the probability that no cows visit it.
We have $n$ fields, so all the cows must choose the other $n-1$
So the probability of a field being visited is $$1-\left(\frac{n-1}n\right)^k$$
Since we have $n$ fields, and we assume each field is equally likely to be chosen, the expected value of the number of fields visited is the sum of the probabilities that each individual field is visited, so all we need to do is multiply our probability by the number of fields, so we have the expected value to be $$n\left( 1-\left(\frac{n-1}n\right)^k \right)$$
As stated in the comments there are many variants of this problem, the generalization I have seen most is distributing balls into bins and finding the expected value of nonempty bins, but cows and fields can be substituted without any issues.
For a sanity check, plugging in $n=k=2$ gives an expected value of $1.5$. This makes sense because there are only two possible ways that cows could go to the two fields, either one field would-be visited or two. There are $4$ ways in total the situation could unfold (cow 1 goes to field 1, cow 2 goes to field 2, both go to field 1, etc). Checking the expected value of visited fields in this particular case could be checked by hand and we also get the answer of $1.5$