Given that $x$ has a Poisson distribution with parameter $\lambda$, which of the following is the correct expression for the probability that $x$ equals $k$?

• A. $\frac{{\lambda }^k}{k}$
• B. $\frac{e^k{\lambda }^k}{k!}$
• C. $\frac{{\lambda }^k}{e^k}$
• D. $\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

Answer: (D)

Verified step by step guidance

1. Recall that a Poisson distribution with parameter \( \lambda \) models the probability of a given number of events \( k \) occurring in a fixed interval of time or space, where these events happen with a known constant mean rate \( \lambda \) and independently of the time since the last event.
2. The probability mass function (PMF) for a Poisson random variable \( X \) is given by the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \( k \) is a non-negative integer (\( k = 0, 1, 2, \ldots \)), \( \lambda > 0 \), and \( e \) is the base of the natural logarithm.
3. Analyze each given expression to see if it matches the PMF formula. The correct expression must have: - \( \lambda^k \) in the numerator, - \( e^{-\lambda} \) (the exponential decay term) in the numerator, - \( k! \) (factorial of \( k \)) in the denominator.
4. Check the terms carefully: the exponential term must be \( e^{-\lambda} \), not \( e^k \) or \( e^{\lambda} \), and the denominator must be \( k! \), not just \( k \) or any other expression.
5. Conclude that the correct expression for \( P(X = k) \) is the one that matches the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]