What is the probability that, when a $6$-sided die is rolled $6$ times, a number greater than $4$ occurs at most $5$ times?
Let $X$ be a random variable that equals $1$ when you roll a $5$ or $6$, and a $0$ when you pick $1$ through $4$. More explicitly speaking:
$$P\left(\sum_{i=1}^6 X_i \leq 5\right)=1-P\left( \sum_{i=1}^6 X_i > 5\right)$$
The specific cases where $\{\sum_{i=1}^6 X_i >5\}$ is when you have all fives or all sixes.
The probability of picking a $5$ six times is $(1/6^6) \cdot {6 \choose 6}$ which equals the probability of picking a six, six times.
I think $$P\left(\sum_{i=1}^6 X_i \leq 5\right)=1-P\left( \sum_{i=1}^6 X_i > 5\right)=1-\frac{2}{6^6}$$
Is this right?
Edit:
I think
$$P\left(\sum_{i=1}^6 X_i \leq 5\right)=1-P\left( \sum_{i=1}^6 X_i > 5\right)= 1- \sum_{j=0}^6 {6 \choose j}(\dfrac{1}{6^6})$$