# Suppose you roll a pair of fair 6-sided dice 36 times. What is the exact probability of getting at least three 9’s?

$\left(\begin{matrix}36 \\ 3\end{matrix}\right) {\left(\frac{1}{4}\right)}^{3} {\left(\frac{3}{4}\right)}^{33} \approx 0.0084$

#### Explanation:

We can find this by using binomial probability:

${\sum}_{k = 0}^{n} {C}_{n , k} {\left(p\right)}^{k} {\left(1 - p\right)}^{n - k} = 1$

Let's look at the rolls possible in rolling two dice:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

There are 4 ways to get a 9 out of 36 possibilities, giving $p = \frac{9}{36} = \frac{1}{4}$.

We roll the dice 36 times, giving $n = 36$.

We're interested in the probability of getting exactly three 9's, which gives $k = 3$

This gives:

$\left(\begin{matrix}36 \\ 3\end{matrix}\right) {\left(\frac{1}{4}\right)}^{3} {\left(\frac{3}{4}\right)}^{33}$

((36!)/(33!3!))(1/4)^3(3/4)^33~~0.0084