Probability that lightbulb stops working in odd year 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 $\frac{F_Y(12)}{F_Y(24)}$.)
