I have a lightbulb that has an exponential lifetime distribution with mean $\mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $\lambda$, and since $E[X] = \frac{1}{\lambda}$,
\begin{align} f(x) &= \frac{1}{\mu}e^{-\frac{1}{\mu}x} \\ F(x) &= 1-e^{-\frac{1}{\mu}x} \end{align}
Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.
I'm having troubles constructing the $P(Y= \mathrm{odd})$ model here, as I don't know how to put together the infinite odd years described above.
There's only one parameter in this exponential model – $\lambda=\frac1\mu$. Thus $$F_Y(y)=1-e^{-\lambda x}$$ $$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^{-12\lambda(k+1)}-1+e^{-12\lambda k}$$ $$=-e^{-12\lambda(k+1)}+e^{-12\lambda k}=e^{-12\lambda k}(1-e^{-12\lambda})$$ Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4\dots$: $$\sum_{n=0}^\infty e^{-24\lambda n}(1-e^{-12\lambda})$$ $$=(1-e^{-12\lambda})\sum_{n=0}^\infty(e^{-24\lambda})^n=\frac{1-e^{-12\lambda}}{1-e^{-24\lambda}}$$ $$=\frac{1-e^{-12/\mu}}{1-e^{-24/\mu}}$$ (We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $\frac{F_Y(12)}{F_Y(24)}$.)