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

Answer: (B)

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. In this formula, \( \lambda^k \) represents the rate parameter raised to the power of the number of events, \( e^{-\lambda} \) accounts for the probability of zero events occurring, and \( k! \) (k factorial) normalizes the probability for the number of ways \( k \) events can occur.
4. Compare the given options to this formula carefully. The correct expression must include the term \( e^{-\lambda} \) (not \( e^k \) or any other exponent), the numerator \( \lambda^k \), and the denominator \( k! \).
5. Therefore, the correct expression for \( P(X = k) \) is the one that matches the formula \( \frac{\lambda^k e^{-\lambda}}{k!} \).