A box contains $10$ blue balls, $7$ red balls, & $8$ green balls. $8$ balls are selected simultaneously from the box. What is the expected number of distinct colors among these 8 balls?
Answer: $$3-\frac{\binom{15}{8} +\binom{18}{8} + \binom{17}{8} }{\binom{25}{8}}$$
Hi sorry I can't comment yet but in @remy's answer, there is a double counting in $P\left(\text {seeing 2 colors}\right)$. The discrepancy is just very small that it doesn't show up in the decimals.
So here is my attempt: \begin{align} & P\left(X=1\right)=\frac{{10\choose8}+{8\choose8}}{25\choose8} =\frac{46}{1081575} \\ & P\left(X=2\right)=\frac{{\sum_{r=1}^{7}{10\choose r}{7\choose 8-r}}+{\sum_{r=1}^{7}{10\choose r}{8\choose 8-r}}+{\sum_{r=1}^{7}{8\choose r}{7\choose 8-r}}}{25\choose8}=\frac{74411}{1081575} \\ & P\left(X=3\right)=\frac{{\sum_{r=1}^{6}{10\choose r}\left({\sum_{k=1}^{7-r}{8\choose k}{7\choose 8-r-k}}\right)}}{25\choose8}=\frac{1007118}{1081575} \\ &\operatorname{E}\left(X\right)=1\cdot\frac{46}{1081575}+2\cdot\frac{74411}{1081575}+3\cdot\frac{1007118}{1081575}=\frac{3170222}{1081575} \end{align} which is consistent with the answer given in the question.
Note: for $P\left(X=3\right)$ the more convenient and exam-friendly way is to use complements but I used cases just to check if there is any mistake in $P\left(X=1\right)$ and $P\left(X=2\right)$.