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

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 $$\alpha=\frac{1-p}{2-p}\ .$$

Let $\alpha$ be the required probability. In order for $X$ to be even, we must have a failure on the first trial, and have the number of trials after the first be odd. The probability of that, given that the first trial was a failure, is $1-\alpha$. Thus $$\alpha=(1-p)(1-\alpha).$$ Solve for $\alpha$.