# Given a fair 12-sided dice: If the dice is rolled ten times, what is the probability of rolling exactly two 6s?

Jun 25, 2018

To three significant figures, the probability is 0.156.

#### Explanation:

The probability of rolling a 6 on any one throw is $\frac{1}{12}$, so the probability of rolling two 6s then the rest of the ten throws not 6s is ${\left(\frac{1}{12}\right)}^{2} \cdot {\left(\frac{11}{12}\right)}^{8} = 0.00346 \ldots$.

Any of the possible lists of throws that have two 6s in total have the same probability as this, so to obtain the probability of rolling two 6s over all the throws, we multiply this probability by the number of possible ways to throw two 6s.

This is a combinatorial question - we are asking how many of the different positions in the ten throws we can put our two 6 throws. This is given by the quantity "10 choose 2", written usually as $\left(\begin{matrix}10 \\ 2\end{matrix}\right)$, and given by the formula ((n),(k))=(n!)/(k!(n-k)!).
These are strongly related to the binomial coefficients, and the Wikipedia article on these makes the connection:
https://en.wikipedia.org/wiki/Binomial_coefficient

((10),(2))=(10!)/(2!8!)=(9*10)/2=45

So there are 45 ways for our ten throws to produce two 6s, and the probability we require is

$45 \cdot {\left(\frac{1}{12}\right)}^{2} \cdot {\left(\frac{11}{12}\right)}^{8} = 0.15579 \ldots$.

So, to three significant figures, the probability is 0.156.