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. 
 A: In general, if $X\sim\operatorname{Geo}(p)$, i.e. $\mathbb P(X=n)=p(1-p)^{n-1}$, $n=1,2,\ldots$, then
\begin{align}
\mathbb E[X] &= \sum_{n=1}^\infty np(1-p)^{n-1}\\
&= -p\sum_{n=1}^\infty \frac{\mathsf d}{\mathsf dp}\left[ (1-p)^n\right]\\
&= -p\frac{\mathsf d}{\mathsf dp}\left[\sum_{n=1}^\infty (1-p)^n\right]\\
&= -p\frac{\mathsf d}{\mathsf dp}\left[\frac {1-p}{p}\right]\\
&= -p\left(-\frac1{p^2} \right)\\
&= \frac1p.
\end{align}
Here we have $p=\frac 5{11}$ so $\mathbb E[X]=\frac{11}5$.
A: 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.
A: $X$ is a geometric random variable with parameter $p$. A way to compute its expected value is through the total expectation theorem:
\begin{align}
E[X] &= E[X\mid X=1]P(X=1) + E[X\mid X>1]P(X>1)\\
\end{align}
When you already know that $X=1$, its expected value is 1, therefore $E[X \mid X=1] = 1$.
When you know that $X>1$, i.e., $X = 2, 3, \ldots$, then you can imagine that you have a "new" random variable, $X-1$, with values $X-1 = 1,2,\dots$, and that is also geometric! This remarkable property of the geometric distribution is formally called the memorylessness property, and implies that $E[X-1 \mid X>1] = E[X]$. Finally,
\begin{align}
E[X] &= E[X\mid X=1]P(X=1) + E[X\mid X>1]P(X>1)\\
     &= P(X=1) + E[(X-1)+1 \mid X>1]P(X>1)\\
     &= p + (E[X-1 \mid X>1]+E[1 \mid X>1])(1-p)\\
     &= p + (E[X]+1)(1-p)\\
\end{align}
This shows that $E[X] = \frac{1}{p}$. In your case $p=\frac{5}{11}$.
Here a great explanation of this approach (minute 24:54) from the MIT's professor John Tsitsiklis: https://youtu.be/-qCEoqpwjf4?list=PLUl4u3cNGP60A3XMwZ5sep719_nh95qOe
Best.
