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.
 A: Using  Binomial Distribution,
$$\binom52\left(\frac16\right)^2\left(1-\frac16\right)^{5-2}$$
A: If you want to use a counting argument, here is how to count the "favourables."
We want to count the number of sequences of length $5$, with exactly two $6$'s.
The places the $6$'s occupy can be chosen in $\binom{5}{2}$ ways.
For each of these ways, the remaining places can be filled with non-$6$'s in $5^3$ ways. 
That gives $\binom{5}{2}5^3$ for the number of favourables. Divide by $6^5$.
Remark: The binomial distribution approach described by lab bhattacharjee is more versatile. 
A: 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
$$ 10 \left(\frac{5}{6}\right)^3 \left(\frac{1}{6}\right)^2 = \frac{625}{3888} $$
