Q:

Find the probability that a geometric random variable $X$ is an even number

Let $\alpha$ be the probability that a geometric random variable $X$ with parameter p is an even number 
a) Find $\alpha$ using the identity $\alpha=\sum_{i=1}^{\infty}P[X=2i]$
b)Find $\alpha$ by conditioning on wether $X=1$ or $X>1$
My attempt for a):
$$\alpha=\sum_{i=1}^{\infty}P[X=2i]=\sum_{i=1}^{\infty}p(1-p)^{2i-1}=p(1-p)\sum_{i=1}^{\infty}(1-p)^{2(i-1)}$$ Letting $j=i-1$
$$p(1-p)\sum_{i=1}^{\infty}(1-p)^{2(i-1)}=p(1-p)\sum_{j=0}^{\infty}(1-p)^{2j}={{p(1-p)}\over {1-(1-p)^2}}$$
hence $$\alpha={{1-p}\over {2-p}}$$
The thing is that I´m having trouble for b): Let $E$ be the event that a geometric random variable $X$ with parameter p is an even number, so : $P(E)=\alpha$ I need to find this probability by conditioning on wether $X=1$ or $X>1$ therefore:
$$P(E)=P(E|X=1)P(X=1)+P(E|X>1)P(X>1)$$
but $P(E|X=1)=0$ hence $P(E)=P(E|X>1)P(X>1)$ then we have that $$P(X>1)=1-P(X\le 1)=1-p$$
But I´m having trouble computing $P(E|X>1)$ Can you help me please? I would really appreciate it :)

A:

We have
$$\eqalign{
  P(\hbox{$X$ is even})
  &=P(\hbox{$X$ is even}\mid X=1)P(X=1)+P(\hbox{$X$ is even}\mid X>1)P(X>1)\cr
  &=P(\hbox{$X$ is even}\mid X>1)P(X>1)\cr}$$
since obviously the first term on the RHS is zero.  Now the probability that $X$ is even, given that the first trial was a failure, is the same as the probability that $X$ is odd (because after the first failure you have an identical experiment, but you need an odd result in this experiment in order to get an even result overall).  So
$$\alpha=(1-\alpha)(1-p)$$
and solving gives
$$\alpha=\frac{1-p}{2-p}\ .$$

