When solving this question

A four sided fair dice is rolled 10 times. What is the probability of two sides being absent and two sides being present?

I get the result $$\frac{{4 \choose 2} \cdot 2^{10}}{4^{10}}$$ yet the official answer is $$\frac{{4 \choose 2}(2^{10}-2)}{4^{10}}$$ Anyone who can explain where this $-2$ comes from?


You have to exclude the possibility that only one side is present. Given that two sides have been excluded, there are two ways this could occur.

  • $\begingroup$ Oh ok now I get it. I kind of feel stupid to not have thought about that. Thank you! $\endgroup$ – ToTom Jul 27 '17 at 10:32

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