Order Statistics of Poisson Distribution I have been given the following question,

Let $n ≥ 2$, and $X_1, X_2, . . . ,X_n$ be independent and identically
  distributed $Poisson (λ)$ random variables for some $λ > 0$. Let
  $X_{(1)} ≤ X_{(2)} ≤ · · · ≤ X_{(n)}$ denote the corresponding order
  statistics.
(a) Show that $P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$.
(b) Evaluate the limit of $P(X_{(2)} > 0)$ as the sample size $n → ∞$.

I tried solving the question on my own and I have also been able to obtain the following expression;
$P(X_{(2)}=0) = 1 - (1-e^{-\lambda})^n-ne^{-\lambda}(1-e^{-\lambda})^{n-1}$
$= 1-(1-e^{-\lambda})^{n-1}(1+e^{-\lambda}(n-1))$ ;
and it can be then shown that
$(1+e^{-\lambda}(n-1)) \le n \quad \text{for all } \lambda > 0 \text{ and } n \ge 2$
and thus
$P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$
but I want to ask that is there any meaning of this statement. I mean is there any significance of the quantity on the left hand side of the inequality so that the inequality can be derived intuitively or by any other method?
 A: $\mathbb{P}(X_{(2)} = 0)$ asked for the probability where the second least r.v. is zero. In other words, it asked for the probability where at least two of $X_1, \cdots X_n$ are zero.
The statement "there are at least two zeros among $X_1, \cdots, X_n$" is false when we have an event where "at least $n-1$ of $X_1, \cdots, X_n$ being greater than zero". The event happens with a probability of at most $$n(1 - e^{-\lambda})^{(n-1)}$$ 
where you have $n$ choices (which $n-1$ r.v.s to choose, or alternatively, which one to leave out), and for each choice you need all $n-1$ of them to be greater than zero, and assume nothing of the one you left out (i.e. adding another multiplier of $1$).
The quoted probability is the upper bound as the event is the least restrictive event among all events that falsify the statement, and any other events will be at least as, or more, restrictive. For example, one can condition on the one variate left out in the selection process, but that involve making more assumptions on the variate's value. Reducing the cardinality of an event (to its subset) will not increase the probability of that event happening, as shown in this Maths.SE question.
To obtain the required probability it is sufficient to exclude the event from the entire event space from consideration. The RHS of the question ($1 - n(1 - e^{-\lambda})^{(n-1)}$) thus forms the lower bound to the required probability.
A: Given: $(X_1, ...,X_n)$ denotes a random sample of size $n$ drawn on $X$, where $X \sim \text{Poisson}(\lambda)$ with pmf $f(x)$:

Then, the pmf of the $2^{\text{nd}}$ order statistic, in a sample of size $n$, is $g(x)$:

... where:


*

*I am using the OrderStat function from the mathStatica package for Mathematica to automate the nitty-gritties, and 

*Beta[z,a,b] denotes the incomplete Beta function $\int _0^z t^{a-1} (1-t)^{b-1} dt$

*Gamma[a,z] is the incomplete gamma function $\int _z^{\infty } t^{a-1} e^{-t} dt$
The exact desired probability $P(X_{(2)}=0)$ is simply: 

The following diagram plots and compares:


*

*the exact solution to $P(X_{(2)}=0)$ just derived (red curve)

*to the bound $P(X_{(2)} = 0) ≥ 1 − n(1 − e^{−λ})^{n−1}$ proposed in the question



... plotted here when $\lambda =3$. 
The bound appears useless for any proper purpose - even a drunk monkey could do better by simply choosing 0 than using the bound proposed that is negative over a huge chunk of the domain (and will get worse as $\lambda$ increases).
