How to get expected value from a probability mass function? A certain biased coin is flipped until it shows heads for the first time. If the probability of getting heads on a given flip is $5/11$ and $X$ is a random variable corresponding to the number of flips it will take to get heads for the first time, the expected value of $X$ is: 
$$E[x] = \sum_{x=1}^\infty{x\frac{5}{11}\frac{6}{11}^{x-1}}$$
I'm not sure how to find an exact value for $E[x]$. I tried thinking about it in terms of a summation of an infinite geometric series but I don't see how that formula can be applied. 

The expectation is not a geometric series (at least not when you write
it directly), but its resemblance to a geometric series is a good observation.
First let's get that factor of $\frac{5}{11}$ out of the way,
because it will become annoying at some point if we keep it inside the
summation.
$$E[x] = \sum_{x=1}^\infty{x\frac{5}{11}\frac{6}{11}^{x-1}}
   = \frac{5}{11} \sum_{x=1}^\infty{x \frac{6}{11}^{x-1}} = \frac{5}{11} S,$$
where
$$
S = \sum_{x=1}^\infty{x \frac{6}{11}^{x-1}}.
$$
Now write out $S$ and $\frac{6}{11}S$:
\begin{align}
\newcommand{x}{\left(\frac{6}{11}\right)}
S    &= 1 \cdot \x^0 + 2 \cdot \x^1 + 3 \cdot \x^2 + 4 \cdot \x^3 + \cdots\\
\frac{6}{11}S &= \phantom{1 \cdot \x^0 + }\ 
                 1 \cdot \x^1 + 2 \cdot \x^2 + 3 \cdot \x^3 + \cdots
\end{align}
From here you should be able to work out what $S - \frac{6}{11}S$ is as a series, taking the difference of the right-hand sides of the two equations above, and then apply what you know about geometric series.
Notice how conveniently $S - \frac{6}{11}S = \frac{5}{11}S$,
which happens to be the value we need in the end.
