Introduction to Probability · Problem 5.1.13

The Poisson distribution models rare events in a fixed interval, and its probability mass function has a clean, exponential form. To find individual probabilities, plug your specific $k$ values directly into $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$—the key is recognizing that all three probabilities must sum to 1, so $P(X > 1) = 1 - P(X = 0) - P(X = 1)$ saves calculation. With $\lambda = 0.3$, you're working with small factorials and a manageable exponential term.

The Poisson distribution models rare events in a fixed interval, and its probability mass function has a clean, exponential form. To find individual probabilities, plug your specific $k$ values directly into $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$—the key is recognizing that all three probabilities must sum to 1, so $P(X > 1) = 1 - P(X = 0) - P(X = 1)$ saves calculation. With $\lambda = 0.3$, you're working with small factorials and a manageable exponential term.