If X is a Poisson variate such that 2 P(X =2) = P(X=1) find P(X = 0)        

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.

Explore how to solve a Poisson distribution problem where 2 P(X = 2) = P(X = 1), leading to finding P(X = 0). This detailed solution covers the Poisson probability mass function and its application. Suitable for advanced mathematics learners.

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