# 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.

• If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on. – Rócherz Mar 15 at 1:21

There's only one parameter in this exponential model – $$\lambda=\frac1\mu$$. Thus $$F_Y(y)=1-e^{-\lambda x}$$ $$P(12k $$=-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)}$$.)
• So $k$ here has to be even right? – PTN Mar 15 at 1:46
• @PTN Yes. ${}{}$ – Parcly Taxel Mar 15 at 1:46