Trouble Understanding Combinatorics Identity for Number of Flips This is my first post on Mathematics StackExchange (and my first time using LaTeX), so I apologize if my question is a little difficult to read.
I am taking an introductory course on probability theory, and came across a problem concerning the number of flips of a coin to reach the desired number of heads:
What is the probability that it takes 10 or more flips of a coin to get three heads?
The probability of n flips is given by $\binom{n-1}{2}$$\left(\frac{1}{2}\right)^n$ , and I seem not to understand exactly why.  I understand that to get the desired answer, you use the complement:
$1-$$\sum_{n=3}^9$$\binom{n-1}{2}$$\left(\frac{1}{2}\right)^n$
However, the initial combination, I do not.
Again, this is my first post, so I am sorry if I did not ask my question correctly.  Any and all help would be much appreciated!  Thanks everyone!

The probability of any single outcome of $9$ coin flips is $(\frac{1}{2})^9$ because for a tail or a head the probability is the same, $\frac{1}{2}$. So the probability of getting $2$ heads is $(\frac{1}{2})^9$ times the number of ways of getting $2$ heads in a series of $9$ flips. This is $^9C_2$ = 36. So the answer is $36\cdot (\frac{1}{2})^9 = .0703125$. We know that if we only get $2$ heads in the first $9$ flips, we are guaranteed to take $10$ or more flips to get $3$ heads.
Update: Getting zero heads with seven flips and one head with eight flips will also guarantee ten or more flips to get three heads.
$P(3H\ge 10) = \ ^9C_2\cdot (\frac{1}{2})^9 + 8\cdot (\frac{1}{2})^8 + (\frac{1}{2})^7$
$= .0703125 + .03125 + .0078125$
$= .109375$
