Let $X_i\sim \mathcal{Poisson}(\lambda)$, where $X_i$ come from a random sample of size $n$ (so they're independent and identically distributed). Let $T=I\lbrace X_1=0 \rbrace$ (indicator function); that is, $T\sim \mathcal{Bernoulli}(e^{-\lambda})$.
Now define $B=\sum_{i=1}^nX_i$. This implies $B\sim\mathcal{Poisson}(n\lambda)$. Using this information, what is the distribution of the conditional random variable:
$$T\mid_{B=b}$$
That is, what is the distribution of $T$ given that $B=b$?
Best Answer
When you have $X_1, ..., X_n \sim \text{IID Pois}(\lambda)$ and you condition on their sum $B = \sum X_i$, the initial values now have a multinomial distribution. Letting $\boldsymbol{X} = (X_1, ..., X_n)$ denote the vector of these original count values, you have:
$$\boldsymbol{X} | B=b \sim \text{Mu} \Bigg( b, \Big( \frac{1}{n}, ..., \frac{1}{n} \Big) \Bigg).$$
From this distribution, the conditional values $X_1, ..., X_n$ have marginal binomial distributions $X_i | B= b \sim \text{Bin} (b, 1 / n)$. Hence, letting $T_i \equiv \mathbb{I}(X_i = 0)$ you therefore have:
$$\mathbb{P}(T_i = 1 | B = b) = \mathbb{P}(X_i = 0 | B = b) = \text{Bin} (0 | b, 1 / n) = \Big( \frac{n-1}{n} \Big)^b .$$