Show that the likelihood that an observation from a Poisson ($\lambda$) distribution takes an odd value (i.e. 1, 3, 5,...) is $\frac{1-{e^-}^{2\lambda}}{2}$.
Since likelihood is given by $L(\lambda)=\prod_{i=0}^n f(X;\lambda)$, if we write probability for each $X=2n+1$ (odd values) i.e. $P(X=1).P(X=3)$ and so on. The Result is not achievable. I mean, what is in the question is Probability or Likelihood function which we calculate to find estimators.
This is actually an interesting computation, and it relies on the hyperbolic functions.
The pdf for a Poisson$(\lambda)$ random variable $X$ is given by:
$$P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$$
and we want to compute the likelihood or probability that $X$ is odd. Any odd integer can be represented as $2n+1$ for some $n \in \mathbb{N}$. So we can use the pdf to write:
$$P(X = 2n+1) = \frac{e^{-\lambda}\lambda^{2n+1}}{(2n+1)!}$$
But we need to do this for every $n$ to get the desired probability so now write it as an infinite sum.
$$P(X \text{ is odd}) = \displaystyle \sum_{n = 0}^{\infty}\frac{e^{-\lambda}\lambda^{2n+1}}{(2n+1)!} =e^{-\lambda} \displaystyle \sum_{n = 0}^{\infty}\frac{\lambda^{2n+1}}{(2n+1)!} $$
Now use the Maclaurin series for the hyperbolic sine 'sinh':
$$\text{sinh } x = \displaystyle \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
$$P(X \text{ is odd}) = e^{-\lambda} \text{sinh }\lambda$$
Now use another clever identity for the hyperbolic sine, namely redefine it in terms of the exponential function:
$$\text{sinh }x = \frac{e^x - e^{-x}}{2}$$
Plug in and solve:
$$P(X \text{ is odd}) = e^{-\lambda} \text{sinh }\lambda = e^{-\lambda}\Big(\frac{e^\lambda - e^{-\lambda}}{2}\Big) = \frac{1 -e^{-2\lambda}}{2} $$