Poisson Distribution 

Poisson Distribution is a type of discrete probability distribution that describes models the number of times an events occurring in a fixed interval/period of time or space, given that these events happen independently and at a constant average rate. It is commonly used in scenarios like counting phone calls, defects, or arrivals over time. Defined by a single parameter λ (mean rate), it helps calculate the probability of observing a specific number of occurrences within the given interval.

1.0What is Meant by Poisson Distribution?

The Poisson Distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, assuming the events occur independently and at a constant average rate. It's widely used in statistics, data science, operations research, and probability theory.

2.0Poisson Distribution Definition

The Poisson Distribution gives the probability of a number of events happening in a fixed time, distance, or space, when these events occur with a known constant rate and independently of the time since the last event. 

3.0Poisson Distribution Formula

The probability of observing exactly x events in an interval is given by:

$P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

Where:

• x: Number of occurrences (0, 1, 2, ...)
• λ: Mean number of occurrences in the interval
• e: Euler's number, approximately 2.71828

4.0Poisson Distribution Graph

The Poisson distribution graph is skewed right for small values of λ and becomes more symmetric as λ increases. It shows the probability mass function with bars centered at whole number values of x.

5.0Poisson Distribution Mean and Variance

• Mean (μ) = λ
• Variance (σ²) = λ

This implies that for a Poisson distribution, mean and variance are equal.

6.0Poisson Distribution Properties

1. Discrete probability distribution.
2. Describes count-based data/events.
3. Applicable when events are independent.
4. Mean = Variance = λ.
5. Only one parameter (λ) governs the distribution.

7.0When to Use Poisson vs Binomial?

Use Poisson Distribution when:

• The number of trials is very large.
• The probability of occurrence per trial is very small.
• You know the average number of events (λ) per interval.

8.0Solved Examples on Poisson Distribution

Example 1: A website gets an average of 2 spam emails per hour. What is the probability that it receives exactly 3 spam emails in an hour?

Solution:
Given λ = 2, x = 3

$P(X=3)=\frac{e^{-2.2}\cdot 2^3}{3!}=\frac{e^{-2.8}\cdot 8}{6}=\frac{0.1353\cdot 8}{6}\approx 0.1804$

Probability = 0.1804

Example 2: On average, 5 patients arrive at a clinic every hour. What is the probability that exactly 7 patients arrive in an hour?

Solution: 

$P(X=7)=\frac{e^{-5}\cdot 5^7}{7!}\approx \frac{0.0067\cdot 78125}{5040}\approx 0.1044$

Probability = 0.1044

Example 3: A call center receives 10 calls per minute on average. Find the probability that it receives no calls in a minute.

Solution: 

$P(X=0)=\frac{e^{-10}\cdot 10^0}{0!}=e^{-10}\approx 4.54\times 10^{-5}$

Probability = 0.0000454

Example 4: A call center receives 5 calls per minute on average. What is the probability that exactly 3 calls are received in a particular minute?

Solution:

Given:

• Λ = 5 (mean rate per minute)
• x = 3

Using the Poisson distribution formula:

$$\begin{gathered} P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!} \\ P(X=3)=\frac{e^{-5}\cdot 5^3}{3!}=\frac{e^{-5}\cdot 125}{6} \\ \text{Using}(e^{-5}\approx 0.0067): \\ P(X=3)=\frac{0.0067\cdot 125}{6}\approx 0.1396 \\ Answer:P(X=3)\approx 0.1396 \end{gathered}$$

Example 5: The average number of errors per page in a printed book is 0.3. What is the probability that a randomly selected page has no errors?

Solution:

Here,
λ = 0.3, x = 0

$$\begin{gathered} P(X=0)=\frac{e^{-0.3}\cdot 0.3^0}{0!}=e^{-0.3}\approx 0.7408 \\ \text{Answer}:(P(X=0)\approx 0.7408) \end{gathered}$$

Example 6: A machine fails on average 2 times per month. What is the probability that it fails at most once in a month?

Solution:

We need P(X ≤ 1) = P(X = 0) + P(X = 1)

Given λ = 2 

$$\begin{gathered} P(X=0)=\frac{e^{-2}\cdot 2^0}{0!}=e^{-2}\approx 0.1353 \\ P(X=1)=\frac{e^{-2}\cdot 2^1}{1!}=2e^{-2}\approx 0.2707 \\ P(X\le 1)=0.1353+0.2707=0.406 \\ \text{Answer:}(P(X\le 1)\approx 0.406) \end{gathered}$$

Example 7: A web server logs 3 errors per hour on average. What is the probability that it logs 4 errors in 2 hours?

Solution:

$$\begin{gathered} \text{In 2 hours}: \\ \lambda =3\times 2=6,x=4 \\ P(X=4)=\frac{e^{-6}\cdot 6^4}{4!}=\frac{e^{-6}\cdot 1296}{24} \\ Using(e^{-6}\approx 0.00248): \\ P(X=4)=\frac{0.00248\cdot 1296}{24}\approx 0.1341 \\ \text{Answer}:(P(X=4)\approx 0.1341) \end{gathered}$$

Example 8: A radioactive source emits on average 1.5 particles per second. What is the probability of detecting at least 2 particles in one second?

Solution:

$$\begin{gathered} \text{We wan}tP(X\ge 2)=1-P(X=0)-P(X=1) \\ Given(\lambda =1.5) \\ (P(X=0)=e^{-1.5}\approx 0.2231) \\ (P(X=1)=\frac{1.5\cdot e^{-1.5}}{1!}=1.5\cdot 0.2231\approx 0.3346) \\ P(X\ge 2)=1-(0.2231+0.3346)=1-0.5577=0.4423 \\ \text{Answer:}(P(X\ge 2)\approx 0.4423) \end{gathered}$$

9.0Poisson Distribution Practice Problems

1. A machine produces 4 defective parts per hour on average. Find the probability of exactly 6 defects in one hour.
2. A customer care center receives 2 complaints per day. What is the probability that it receives 5 complaints in a day?
3. Emails arrive at a rate of 3 per minute. What is the probability that no emails arrive in the next 2 minutes?
4. A power outage occurs on average once per month in a city. What is the probability of exactly 2 outages in a month?
5. What is the probability of at most 2 accidents in a week if the average rate is 1.5 accidents per week?