Probability of dice being rolled a $7$ eight times in $10$ rolls

Question

A pair of dice is rolled, and the player watches for sevens. What is the probability that it takes exactly $10$ rolls to observe $8$ sevens?

My approach is that I know the dice are rolled until $8$ sevens occur. I know the probability of rolling a $7$ is $\frac{1}{6}$.
I want to know how to find the probablity of obtaining $7$ sevens in the first $9$ rolls followed by a seven on the tenth roll? This is what I need to do to answer the question.

Answer 1

Hint: On each roll (of the two dice), you can either have a success (a seven) or a failure (anything else). Additionally, the order of the successes doesn't matter, as long as you get eight out of ten.

Do you know any distribution that would be suited for this?

Answer: The answer is to use the binomial distribution, which gives the probability of $k$ successes out of $n$ trials, when the probability of a success is $p$, given by $$\binom{n}{k}p^k(1-p)^{n-k}.$$ In your case the probability of a succes is $p=\frac{1}{6}$, the number of trials is $10$ and the number of successes for which we want to calculate the probability is $k=8.$ Insert these values into the formula and get the desired result.

Edit: As mathreadler correctly points out, the problem formulation actually asks for the probability of getting $k=7$ successes in $n=9$ trials and then getting a seven on the tenth trial. With these values of $n$ and $k$, the probability is then $$\binom{n}{k}p^{k+1}(1-p)^{n-k}.$$