So, I've been trying to teach myself probability theory, and I think I have the idea down, but I wanted to make sure I am on the right track. Thus, I was wondering if people here could confirm that I am doing this right...
If I have an urn with k balls all colored distinctly, I want to calculate the probability that choosing k+2 of them(with replacement) will give at least one ball of each color.
I think this will be $$1-\sum_{i=1}^{k-1}{k\choose i}(\frac{k-i}k)^{k+2}(-1)^{i+1}$$ but I am not sure. If someone can either confirm this or let me know what I am doing wrong, that would be great.
Your Inclusion/Exclusion expression is correct, and works equally well for $m\ge k$ trials. We consider a more simple-minded cases approach that works well for $m=k+2$, but that is unwieldy for larger $m$.
There are $k^{k+2}$ equally likely sequences of $k+2$ colours. We count the favourables, in which all $k$ colours are represented.
Let $k\ge 2$. There are two types of favourable: (i) one colour occurs three times, and the other $k-1$ colours appear once each, and (ii) two colours appear twice each, and the other $k-2$ colours appear once each.
First we count the Type (1) favourables. The colour that appears three times can be chosen in $\binom{k}{1}$ ways. The places it occurs can be chosen in $\binom{k+2}{3}$ ways. And the rest of the colours can occupy the remaining $k-1$ places in $(k-1)!$ ways, for a total of $\binom{k}{1}\binom{k+2}{3}(k-1)!$ ways.
We next count the Type (ii) favourables. The colours that occur twice can be chosen in $\bino{k}{2}$ ways, and placed in $\binom{k+2}{2}\binom{k}{2}$ ways. Then the rest can be filled in $(k-2)!$ ways, for a total of $\binom{k}{2}\binom{k+2}{2}\binom{k}{2}(k-2)!$.
Remark: It is easy to miscount the Type (ii) cases, and get an expression twice the correct one.