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

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 for \( P(X = k) \), which involves \( \lambda \), \( k \), and the exponential function.
3. Write down the general formula for the Poisson PMF: \[ P(X = k) = \frac{\lambda^{k} e^{-\lambda}}{k!} \] where: - \( \lambda^{k} \) represents the rate parameter raised to the power of the number of events, - \( e^{-\lambda} \) is the exponential decay factor, - \( k! \) is the factorial of \( k \), accounting for the number of ways the events can occur.
4. Compare the given options to this formula carefully, paying attention to the placement of the negative sign in the exponent and the factorial in the denominator.
5. Identify the correct expression as the one that exactly matches the formula \( \frac{\lambda^{k} e^{-\lambda}}{k!} \), ensuring the exponent on \( e \) is \( -\lambda \) and not \( k \), and that the factorial is in the denominator.