What is the probability of rolling exactly 3 6s and 3 5s in 6 rolls? Not sure how to approach this question, I know how to do it for 3 6s in 6 rolls, but not with the added condition of the 3 5s. Any help would be appreciated.
 A: I assume, that you´re rolling one fair six-sided dice six times. There are $6^6=b$ possible outcomes. And the number of desired outcomes can be calculated by the formula $\frac{n!}{(n-m)!\cdot m! }=a$. $n!$ is the factorial of n.
n is the number of rollings. $m$ is the number of rollings with the outcome 5. $n-m$ is the number of rollings with the outcome 6.
The probability of 3 5´s and 3 6´s in 6 rolls then is $\frac{a}{b}$
A: Hints:
If you induce an ordering of the rolls then probability of reaching $555666$ is (if the die is $6$-sided and fair) just $\left(\frac16\right)^6$, right?
But there are more possibilities (e.g. $556566$). How many?...
These events exclude each other so you allowed to add the corresponding probabilities.
Fortunately all these possibilities have equal probability to occur.
A: The good events are $\binom{6}{3}$. The possible ones are $6^6$. Therefore
$$
\frac{\binom{6}{3}}{6^6}=\frac{5}{2^4\cdot 3^6}
$$
A: Like N.F. Taussig already mentioned the number of orders is relevant, that means for example: 6, 6, 6, 5, 5, 5 and 6, 6, 5, 6, 5, 5 and so on...
All these combinations come with the same probability of $(1/6)^6$ and has to be added.
