A hunter killed 10 animals, 2 of which are prohibited from being hunted. A wildlife inspector checks 2 random animals out of the total 10. What's the probability that the inspector will not check the 2 prohibited animals?
The answer is $$ \frac{\binom{8}{2}}{\binom{10}{2}} = \frac{28}{45} $$
I wanted to solve the problem via calculating the probability of the event that the inspector did check the 2 prohibited animals (let this event be $A$). Then the event that the inspector doesn't check the 2 prohibited animals would be $A^C$, $A$'s complement. $$ A = \frac{\binom{2}{2}}{\binom{10}{2}} = \frac{1}{45} $$ Then: $$ A^C = 1 - \frac{1}{45} = \frac{44}{45} $$
Which frankly I guess doesn't make sense but I don't understand why.
The ways to inspect a prohibited animal would be $${2\choose2}+{2\choose1}{8\choose1}$$ Because you could either choose both prohibited animals or one prohibited animal and one legal animal.