Given that $X$ is a Poisson random variable, we can write the probability mass function (PMF) of $X$ as:

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

where $\lambda$ is the rate parameter of the Poisson distribution.

We are given the equation:

$2 \cdot P(X = 2) = P(X = 1)$

Step 1: Express $P(X = 2)$ and $P(X = 1)$ using the PMF

$P(X = 2) = \frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^2 e^{-\lambda}}{2}$

$P(X = 1) = \frac{\lambda^1 e^{-\lambda}}{1!} = \lambda e^{-\lambda}$

Step 2: Substitute these into the given equation

$2 \cdot \frac{\lambda^2 e^{-\lambda}}{2} = \lambda e^{-\lambda}$

Simplifying, we get:

$\lambda^2 e^{-\lambda} = \lambda e^{-\lambda}$

Step 3: Solve for $\lambda$

Since $e^{-\lambda}$ is common on both sides, we can divide both sides by $e^{-\lambda}$, assuming $\lambda \neq 0$:

$\lambda^2 = \lambda$

Dividing both sides by $\lambda$ (assuming $\lambda \neq 0$):

$\lambda = 1$

So, the parameter $\lambda = 1$.

Step 4: Find $P(X = 0)$

Now, we calculate $P(X = 0)$ using the PMF:

$P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = \frac{e^{-\lambda}}{1} = e^{-1}$

Thus, the probability $P(X = 0) = e^{-1}$.

Final Answer

$P(X = 0) = e^{-1} \approx 0.3679$

Would you like more details or have any other questions?

Here are some related questions:

1. How do you derive the mean and variance of a Poisson distribution?
2. What are some practical examples of phenomena that follow a Poisson distribution?
3. How does the Poisson distribution relate to the exponential distribution?
4. What is the significance of the rate parameter $\lambda$ in the Poisson distribution?
5. How would the calculation change if $P(X = 3) = P(X = 2)$?

Tip: The Poisson distribution is often used to model the number of events occurring within a fixed interval of time or space, particularly when events happen independently of each other.