<s>[INST] Probably of 2 six in 5 dice rolls What is the probability of obtaining exatcly 2 six when rolling a dice 5 times?
In order to obtain this probability, I will need to devide the number of favorable events by the number of possible events. I believe the denominator is $6^5$. But I am having troubles figuring out the numerator.
 [/INST] The generating function for the five throws is 
$$ \left(\frac{s}{6}+\frac{5 o}{6}\right)^5 $$ where o are the other faces (1,2,3,4,5) and s represents the six. You need the coefficient of $$ s^2o^3 $$. This you get from Pascals triangle which tells us that the expansion of $$(a+b)^5$$ is
$$(a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 $$
So $$ a = \frac{5}{6} $$ and $$ b = \frac{1}{6} $$ gives us using the third term
$$ 10 \left(\frac{5}{6}\right)^3 \left(\frac{1}{6}\right)^2 = \frac{625}{3888} $$
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