How to find the mean of a biased coin's probability distribution? I asked a question on this site about the same scenario:
The probability of a head occurring when a biased coin is tossed is $p$, where $p < 1$. Let the random variable $X$ represent the number of tosses up to and including the first toss on which a tail occurs.
So I found out that the probability is $p^{n-1} (1-p)$
I now need to find out the $E(X)$ is (which I think is the same as the mean).
I tried tabulating the values and adding them but got nowhere. How do I get the $E(X)$ value? 
The answer is meant to be $\frac{1}{1-p}$.
 A: A bit more machinery than perhaps required. Using the law of total expectation we have that
$$
\begin{align}
EX&=E(X\mid X>1)P(X>1)+E(X\mid X=1)P(X=1)\\
&=(1+EX)p+1(1-p)\tag{0}\\
&=1+pEX
\end{align}
$$
where in (0) we used the fact that $P(X=1)=1-p$ (tails on the first try), $E(X\mid X=1)=E(1)=1$ and that $P(X>1)=p$ since the event $(X>1)$ corresponds to heads on the first toss. Finally $E(X\mid X>1)=1+EX$ since we failed to get a tail on the first toss and the process starts anew thereafter. Hence
$$
EX=\frac{1}{1-p}.
$$ 
A: Hint: To compute $\sum_{n=1}^\infty np^{n-1}$, try to use the fact:
$$(\sum_{n=1}^\infty x^n)’=\sum_{n=1}^\infty nx^{n-1},$$
for $x\in [0,1)$.
A: $$P(X = k) = p^{k-1}\times (1-p)$$
This is a Hypergeometric Distribution.
$$E(X) = \sum_{k=1}^{\infty} k \times P(X=k)$$
$$\implies E(X) =  \sum_{k=1}^{\infty} k \times p^{k-1}\times (1-p) \quad (1)$$
$$p\times E(X) = \sum_{k=1}^{\infty} k \times p^{k}\times (1-p) \quad (2)$$
Now do $(1)-(2)$. 
$$(1-p) E(X) = (1-p)\times \left[\sum_{k=0}^{\infty} p^k\right]$$
$$\implies E(X) =\frac{1}{1-p}$$
