Discrete random variable. Tossing a coin. Two coins are simultaneously tossed until one of them comes up a head
and the other a tail. The first coin comes up a head with probability $p$ and the second with probability $q$. All tosses are assumed independent. 

(a) Find the PMF, the expected value, and the variance of the number of tosses.

$$P(X=k) = (1-p(1-q) - q(1-p))^{k-1}(p(1 − q) + q(1 − p)), \ \ k=1,2...$$
And the above is clear for me.
Now, we would like determine expected value:
$$E[X] = \sum_{k&gt;0} kP(X=k)$$
and from solution:
$$E[X] = \frac{1}{p(1-q)+q(1-p) }$$
And I don't understand this. Please explain me.
Hint. One may recall the useful evaluation:
$$
1+r+r^2+\cdots+r^n+...=\frac{1}{1-r}, \quad |r|&lt;1, \tag1
$$ giving by differentiation and multiplication by $r$:

$$
\sum_{k&gt;0} k\: r^k=r+2r^2+3r^3+\cdots+nr^{n}+\cdots=\frac{r}{(1-r)^2}, \quad |r|&lt;1\tag2
$$ 
and now put
$$
r:=1-p(1-q) - q(1-p)
$$ in $(2)$.
I'm not great with anything other than basic math, but I can tell you one thing:  The probabibility of the next coin tossed coming up heads is 50% and tails 50%.  You don't even need math for that one :).  That probability will remain the same the next million tosses.