If $X_1, X_2, \ldots, X_n$ are a random sample from a Poisson Distribution with mean $\vartheta>0$, how do you find $P(X\le 1)$ in terms of $\vartheta$?
I've proven that summing $X_i$ for $i=1,\ldots,n$ is sufficient for $\vartheta$, which I expected would be helpful, but I've got no clue where to take it from here. Not even sure that was a productive step, but it's helped things fall into place before.
Thanks guys!
If by $X$ you mean $X_i$ then the answer is just $$ P[X_i \leq 1] = P[X_i = 0] + P[X_i = 1] = e^{-\theta}(1 + \theta). $$ If by $X$ you mean $X = \sum_{i=1}^n X_i$, then since $X_i$ $i=1,...,n$ is a random sample, $X$ is a Poisson($n\theta$) r.v. and so $$ P[X \leq 1] = P[X = 0] + P[X = 1] = e^{-n \theta}(1 + n\theta). $$ No one can give a definite answer until you clarify what $X$ is though.