To solve this problem, we must first clarify the context of the question related to the Poisson distribution.

The Poisson distribution is defined by the probability mass function:

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

where:

• $\lambda$ is the average rate of occurrence (mean),
• $k$ is the specific number of occurrences,
• $P(X = k)$ is the probability of $k$ occurrences.

Given:

1. $P_1 = P_2$, meaning the probabilities of $X = 1$ and $X = 2$ are equal.
2. We are to find $P_4$, the probability of $X = 4$.

Step 1: Solve for $\lambda$ using $P_1 = P_2$:

From the probability mass function, substitute $k = 1$ and $k = 2$:

$P_1 = \frac{\lambda^1 e^{-\lambda}}{1!}, \quad P_2 = \frac{\lambda^2 e^{-\lambda}}{2!}.$

Since $P_1 = P_2$: $\frac{\lambda e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}.$

Canceling $e^{-\lambda}$ and $\lambda$ (assuming $\lambda > 0$): $1 = \frac{\lambda}{2}.$

Solve for $\lambda$: $\lambda = 2.$

Step 2: Find $P_4$:

Substitute $\lambda = 2$ and $k = 4$ into the probability mass function: $P_4 = \frac{\lambda^4 e^{-\lambda}}{4!}.$

$P_4 = \frac{2^4 e^{-2}}{4!}.$

$P_4 = \frac{16 e^{-2}}{24}.$

Simplify: $P_4 = \frac{2}{3} e^{-2}.$

Final Answer:

$P_4 = \frac{2}{3} e^{-2}.$

Let me know if you'd like a numerical approximation or more clarification!

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Related Questions:

1. What is the general formula for the Poisson distribution probabilities?
2. How is $P_3$ calculated for this problem?
3. What are the properties of the Poisson distribution?
4. How does the Poisson distribution differ from the binomial distribution?
5. Can $P_k$ be maximized for specific values of $k$ and $\lambda$?

Tip:

The Poisson distribution is most useful for modeling events that occur independently and at a constant average rate.