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

Answer: (C)

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 form of 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. The correct expression must include the term \( e^{-\lambda} \) (the exponential with a negative exponent), which reflects the probability of no events occurring in the complementary intervals.
5. Confirm that the numerator contains \( \lambda^{k} \) multiplied by \( e^{-\lambda} \), and the denominator is \( k! \). This matches the standard Poisson PMF and thus is the correct expression for \( P(X = k) \).