Let $X$ be an exponential random variable with parameter $\lambda > 0$. Calculate the probability that $X$ belongs to the interval $[n,n+1]$ with $n$ odd.
From the solution (Q1) I see we need to sum over all odd $n$ but I'm confused as to why the answer doesn't depend on $n$? Surely as $n \rightarrow \infty $ the probability falls to $0$ and is different than for small $n$?
Any help on my misunderstanding appreciated.
The solution you link to is way too complicated. I'm surprised that this is the solution suggested in an MIT course. It teaches you to perform a mechanical calculation instead of applying an idea.
The map $x\to x+1$ is a bijection from the non-negative reals with even integer part to those with odd integer part. The probability density at $x+1$ is $\mathrm e^{-\lambda}$ times the density at $x$. It follows that the integral of the density over the non-negative reals with odd integer part is $\mathrm e^{-\lambda}$ times the integral over those with even integer part. Thus, if we denote the probability to obtain an even integer part by $p_\text e$ and the probability to obtain an odd integer part by $p_\text o$, we have $p_\text o=\mathrm e^{-\lambda}p_\text e$. Since $p_\text o+p_\text e=1$, it follows that
\begin{eqnarray*} p_\text e&=&\frac1{1+\mathrm e^{-\lambda}}\;,\\ p_\text o&=&\frac{\mathrm e^{-\lambda}}{1+\mathrm e^{-\lambda}}\;.\\ \end{eqnarray*}
(As clarified in the comments, there's no dependence on $n$ because the intended interpretation is “any interval” instead of “the interval”).