TITLE: Coin toss problem, get exactly 2 heads in 5 tosses
QUESTION [6 upvotes]: Suppose we toss a fair coin until we get exactly 2 heads. What is
  the probability that exactly 5 tosses are required?


My try: 
We have to make sure that the first 4 tosses does not have 2 heads and the last toss must be a head. That is, the first 4 tosses need to contain 1 head and 3 tails. The probability of this event is $\frac{4}{2^4}=1/4$. Then the probability of 5th toss is head is $1/2$. Hence, in the end the answer is $\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$.
Am I correct?

REPLY [3 votes]: This really sounds like you are saying the following:
"
The probability  of the number X of Bernoulli trials needed to get n successes
"
If that is the case we can just apply the negative binomial formula:
\begin{equation}
C^{n + r -1}_{r-1} p^{r} (1-p)^{n}
\end{equation}
$r$ is the number of successes and $n$ is the number of failures so $r=2, n=3$ which yields:

$n+r-1= 2+3 -1=4$
$C^{n+r-1}_{r-1} = C^{4}_{1} = 4=6$

\begin{equation}
C^{4}_{1} (0.5)^{2} (0.5)^{3}
\end{equation}
\begin{equation}
4(0.5)^{2} (0.5)^{3} = \frac{4}{2^{5}} = \frac{1}{8}
\end{equation}
Which agrees with your answers sorry for the typos :)