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?
Best Answer
You're nearly right. You've found the probability that the first three are 6 and the remaining seven are not 6. For example, you haven't taken account of $6111616111$.
Now, there are $\binom{10}{3}$ possible ways we could have got three 6s, so you need to multiply your answer by $\binom{10}{3}$.