Flip a fair coin repeatedly. the probability that the number of coin flips until the second head occurs is at most 10

Question

Flip a fair coin repeatedly. What is the probability that the number of coin flips until the second head occurs is at most 10?

Is this a negative binomial distribution?
$P(X \le 10)= \binom{10}{2} \cdot (\cfrac{1}{2})^2 \cdot (\cfrac{1}{2})^8$

Answer 1

Here's the way I would do it:

Consider the probability that $0$ of the first $10$ flips are heads.

That happens with $\displaystyle \left (\frac{1}{2}\right)^{10}=\frac{1}{1024}$ probability.

Now consider the probability that $1$ of the first $10$ flips are heads.

That happens with $\displaystyle \binom{10}{1}\cdot\left (\frac{1}{2}\right)^{1}\cdot\left (\frac{1}{2}\right)^{9}=\frac{10}{1024}$ probability.

The complement of those two cases is that $2+$ of the first $10$ flips were heads, which implies it took at MOST $10$ flips for the second head to be flipped.

So your answer should be $\displaystyle \frac{1013}{1024}$.