What is the chance of rolling a die and getting the number six three times at exactly 10 rolls? 
What is the chance of rolling a die and getting the number six three times at exactly 10 rolls?

I was asked this question in my statistic class. I thought the way to do this was $(1/6)^3 \times (5/6)^7$, because that is getting six 3 times and not getting it 7 times. However, that's wrong, I figured that it's because that $(1/6)^3$ would be getting 6 three times in a row. 
Could you explain how to do this?

You’ve computed the probability of rolling six three times and some other number seven times in that order. You also need to account for all of the other ways in which three sixes can come up in ten rolls. The binomial distribution is called for here. If we let $X$ be the number of times a six is rolled, then $$
P(X=3)={10\choose 3}\left(\frac16\right)^3\left(\frac56\right)^7.
$$ As you can see, there’s an extra factor of ${10\choose3}$ that you missed.
