Markov’s inequality and Poisson distribution

Question

Let X be a random variable having the Poisson distribution with parameter 1. What does Markov’s inequality (p.72) imply about the probability P{X ≥ 2}?

Answer 1

Markov's Inequality states that if $X$ is a nonnegative random variable and $c > 0$, then $$P(X \geq c) \leq \frac{\mathbb{E}[X]}{c}.$$

We have $X \sim \text{Poisson}(1)$, meaning that $\mathbb{E}[X] = \lambda = 1$. We can set $c = 2$. Then, we obtain the bound

$$P(X \geq 2) \leq \frac{\mathbb{E}[X]}{2} = \frac{1}{2}.$$

Therefore, we conclude $P(X \geq 2)$ will be at most $0.5$.