I throw a pot of $10$ dice. What is the probability that I get at least two sixes?

Question

Problem: I throw a pot of $10$ dice. What is the probability that I get at least two sixes?

My solution: $1 - \left(\frac{5}{6}\right)^{10} - \frac{1}{6}\left(\frac{5}{6}\right)^9 \cdot 10$. I got this by doing the following.

The probability of getting no six is $\left(\frac{5}{6}\right)^{10}$.

The probability of getting exactly one six is $\frac{1}{6}\left(\frac{5}{6}\right)^9 \cdot 10$.

Hence, the probability that I get at least two sixes is $1 - \left(P[\text{no six}] + P[\text{exactly one six}]\right)$.

Question: Am I right? This seems more efficient than figuring out the probability of getting two sixes, three sixes, and so on, and then adding up those probabilities.

Answer 1

This is very slightly wrong. The probability of getting exactly one six is actually $\frac{1}{6}\left(\frac{5}{6}\right)^9 \cdot 10$.

Answer 2

We calculate this answer using the binomial distribution. The probability of getting $n$ sixes out of 10 is $$ P(n)={10\choose n} \left(1\over6\right)^n \left(5\over6\right)^{10-n} $$ so in this case the probability of zero or one sixes is $$ \begin{align} P(n <2) &={10\choose 0} \left(1\over6\right)^0 \left(5\over6\right)^{10} +{10\choose 1} \left(1\over6\right)^1 \left(5\over6\right)^{10-1}\\ &= \left(5\over6\right)^{10} +10 \left(1\over6\right) \left(5\over6\right)^{9}\\ \end{align} $$