What is the probability of rolling at least two 6's on twelve rolls of a fair 6-sided die.
I am using the complement to solve the question
$$1-\frac {5^{12}+(_{12}C_ 1)5^{11}}{6^{12}}$$
$5^{12}$ is the probability of not rolling a six
$6^{12}$ is the sample space
My question is why does the binomial coefficient have to be multiplied $5^{11}$ where the binomial coefficient is a "6" on the die.
This method I find difficult but intriguing.
You are using a counting argument (I would probably use the equivalent Binomial distribution argument.)
You counted correctly the strings that have no $6's$. We now count the strings of length $12$ that have exactly one $6$.
Where is the lone $6$ in the string? The place can be chosen in $\dbinom{12}{1}$ ways.
For each way of deciding where the $6$ goes, there are $11$ slots left, to be filled with non-$6$'s.
At every empty slot, we have $5$ choices of what to put into it. So given the locatin of the $6$, the remaining slots can be filled in $5^{11}$ ways, for a total of $\dbinom{12}{1}5^{11}$.
$\left(\frac56\right)^{12}$ is the probability that no sixes are rolled in 12 rolls.
$\frac16\left(\frac56\right)^{11}$ is the probability that a six is rolled on the first roll and then none are rolled in the remaining $11$ rolls. Now we have to count the possibilities that each of the $12$ dice could be six and the rest not. Therefore, we multiply by $\binom{12}{1}=12$.
Thus, the probability of rolling at least $2$ sixes would be the complement of the sum of these probabilities: $$ 1-\left(\frac56\right)^{12}-\binom{12}{1}\frac16\left(\frac56\right)^{11} $$
If we want to count the number of ways to roll exactly $k$ sixes, this would be $\binom{12}{k}\left(\frac16\right)^k\left(\frac56\right)^{11-k}$ since $\binom{12}{k}$ is the number of ways to choose the dice on which the sixes are rolled.
For the $\binom{12}{1}\frac{5^{11}}{6^{12}}$ part, you are finding the probability that exactly one roll gives a 6, out of the $6^{12}$ outcomes. Therefore, there are $\binom{12}{1}$ rolls that can give a 6, and the other $12-\binom{12}{1}$ rolls can be ranging from 1 to 5. Hence the $5^{11}$.
