binomial distribution - If you throw a fair dice $10$ times, what is the probability to throw number $6$ at most once?

If you throw a fair die $10$ times, what is the probability to throw number $6$ at most once?

I thought the answer was the sum of probability to throw $6$ once in $10$ throws plus probability to throw $6$ zero times in $10$ throws:
$$\frac{1}{6}\left(\frac{5}{6}\right)^9+\left(\frac{5}{6}\right)^{10}$$
Why is this not correct?

Answer

You forgot the binomial coefficient. It should be

$$\binom{10}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{10}+\binom{10}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^9 = 0.4845167$$

In other words, you need to count which spots get a six. In the first case, zero spots get a six and there are $\binom{10}0$ ways to do that. In the second case, you need to get one six and there are $\binom{10}{1}$ ways to choose the spot where six lands.