To find the probability that $X$ is at least 2, where $X$ follows a Poisson distribution with parameter $\lambda = 2.0$, we want to calculate $P(X \geq 2)$.

The Poisson distribution probability mass function (PMF) is given by:

$P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}$

To find $P(X \geq 2)$, we can use the complement rule:

$P(X \geq 2) = 1 - P(X < 2)$

where

$P(X < 2) = P(X = 0) + P(X = 1)$

Let's calculate these probabilities and then find $P(X \geq 2)$.The probability that $X$ is at least 2, given that $X$ follows a Poisson distribution with $\lambda = 2.0$, is approximately 0.594.

Would you like more details on this calculation or have any other questions?

Here are 5 related questions that might interest you:

1. How does the value of $\lambda$ affect the shape of a Poisson distribution?
2. What is the probability that $X$ is exactly 2 in this distribution?
3. How would you calculate $P(X \leq 2)$ for the same distribution?
4. What is the expected value and variance of a Poisson distribution?
5. How can Poisson distributions be used to model real-world events?

Tip: When working with Poisson distributions, the complement rule is often a useful shortcut to calculate cumulative probabilities like $P(X \geq k)$.