The Poisson distribution is a statistical tool crucial in predicting the probability of a given number of events occurring within a defined period. It's particularly useful when events happen independently and at a constant average rate.

Understanding the Poisson Distribution

Basics

The Poisson distribution shows how likely a number of events will happen in a set period of time or space, assuming these events occur with a fixed average rate and independently of the time since the last event.

Key Points

• Independence: Events don't affect each other.
• Constant Rate: Events happen at a steady average rate.
• Formula: The chance of seeing $k$ events is $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average number of events, and $k$ is the event count.

When to Use

• For random, independent events happen at a constant rate.
• Ideal for rare events over many trials.

Examples

Example 1: Helpdesk Calls

• Scenario: A helpdesk gets 5 calls per hour on average.
• Question: What's the chance of exactly 7 calls in an hour?
• Given: $\lambda = 5, k = 7$.
• Formula: $P(X = 7) = \frac{e^{-5} \cdot 5^7}{7!}$
• Steps:
  • $e^{-5} \approx 0.0067$ (since $e \approx 2.71828$)
  • $5^7 = 78125$
  • $7! = 5040$
  • Thus, $P(X = 7) \approx \frac{0.0067 \times 78125}{5040} \approx 0.1044$
• Result: The probability of exactly 7 calls is about 10.44%.

Example 2: Radioactive Decay

• Scenario: A substance decays at a rate of 20 particles per minute.
• Question: What's the probability of 25 decays in one minute?
• Given: $\lambda = 20, k = 25$.
• Formula: $P(X = 25) = \frac{e^{-20} \cdot 20^{25}}{25!}$
• Steps:
  • $e^{-20} \approx 2.06 \times 10^{-9}$
  • $20^{25}$ is a large number, about $3.36 \times 10^{34}$
  • $25! = 1.55 \times 10^{25}$
  • Thus, $P(X = 25) \approx \frac{2.06 \times 10^{-9} \times 3.36 \times 10^{34}}{1.55 \times 10^{25}} \approx 0.0446$
• Result: The probability of exactly 25 decays occurring in one minute is approximately 4.46%.