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AP Statistics & IB Math AA HL — Exam Mastery Series

Poisson Distribution

Concept review, worked examples, and 20 exam-style problems with full solutions

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Concept Review
Poisson Distribution

Definition

A discrete random variable \(X\) follows a Poisson distribution if it counts the number of events occurring in a fixed interval (time, space, or volume), where events occur independently at a constant average rate \(\lambda\).

We write:   \(X \sim \text{Po}(\lambda)\)

Probability Mass Function
\[ P(X = k) = \frac{e^{-\lambda} \lambda^{k}}{k!}, \quad k = 0, 1, 2, 3, \ldots \]

where \(\lambda > 0\) is the mean number of events per interval, \(e \approx 2.71828\), and \(k! = k \times (k-1) \times \cdots \times 1\).

Key Properties

Mean \(E[X] = \lambda\)
Variance \(\text{Var}(X) = \lambda\)
Std Deviation \(\sigma = \sqrt{\lambda}\)
Notation \(X \sim \text{Po}(\lambda)\)

Rate scaling: If \(X \sim \text{Po}(\lambda)\) per unit time, then over \(t\) units: \(X_t \sim \text{Po}(\lambda t)\).

Cumulative Probability
\[ P(X \leq k) = \sum_{i=0}^{k} \frac{e^{-\lambda}\lambda^{i}}{i!} \]

Four Conditions for Poisson Model

1. Discrete Events Events are countable and occur one at a time (not simultaneously).
2. Independence One event does not affect the probability of another.
3. Constant Rate The average rate \(\lambda\) is constant across the interval.
4. Fixed Interval Events are counted over a fixed period, area, or volume.

Poisson vs. Binomial Approximation

When \(n\) is large and \(p\) is small (\(n \geq 50,\ np \leq 5\)), the Binomial can be approximated by Poisson with \(\lambda = np\).

Feature Binomial \(B(n,p)\) Poisson \(\text{Po}(\lambda)\)
Trials Fixed \(n\) Not fixed
Outcomes Success / Failure Count of events
Parameter \(n, p\) \(\lambda\)
Mean \(np\) \(\lambda\)
Variance \(np(1-p)\) \(\lambda\)
Use when Finite trials, moderate \(p\) Rare events, large \(n\), small \(p\)
Quick Reference
Formulas to Memorize
Formula 01
PMF
\(P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}\)
Formula 02
Mean & Variance
\(E[X]=\text{Var}(X)=\lambda\)
Formula 03
Rate Scaling
\(X\sim\text{Po}(\lambda) \Rightarrow X_t\sim\text{Po}(\lambda t)\)
Formula 04
Complement Rule
\(P(X \geq 1)=1-e^{-\lambda}\)
Formula 05
Binomial Approx.
If \(n\geq50,\ np\leq5\): use \(\lambda=np\)
Formula 06
Sum of Poissons
If \(X\sim\text{Po}(\alpha),\ Y\sim\text{Po}(\beta)\) independent:
\(X+Y\sim\text{Po}(\alpha+\beta)\)
Worked Examples
See It First
EX 01 Direct PMF Calculation
A hospital emergency room receives an average of 4 patients per hour. Assuming the number of arrivals follows a Poisson distribution, find \(P(X = 3)\).
  • 1
    Identify \(\lambda = 4\) (average rate per hour), \(k = 3\).
  • 2
    Apply the PMF:  \(P(X=3)=\dfrac{e^{-4}\cdot 4^3}{3!}\)
  • 3
    Calculate: \(4^3 = 64,\quad 3! = 6,\quad e^{-4} \approx 0.01832\)
  • 4
    \(P(X=3)=\dfrac{0.01832 \times 64}{6}=\dfrac{1.1725}{6}\approx 0.1954\)
Answer: P(X = 3) ≈ 0.1954
EX 02 Rate Scaling over Different Interval
A radioactive source emits on average 6 particles per minute. Find the probability that exactly 2 particles are emitted in a 30-second interval.
  • 1
    Original rate: \(\lambda = 6\) per minute. New interval: 30 sec = 0.5 min.
  • 2
    Scale rate: \(\lambda' = 6 \times 0.5 = 3\) particles per 30 seconds.
  • 3
    \(P(X=2)=\dfrac{e^{-3}\cdot 3^2}{2!}=\dfrac{0.04979 \times 9}{2}\approx 0.2240\)
Answer: P(X = 2) ≈ 0.2240
EX 03 Cumulative Probability
Defects in a fabric roll occur at an average rate of 2 per metre. Find the probability that a 1-metre section has at most 3 defects.
  • 1
    \(\lambda = 2\), find \(P(X \leq 3) = P(0)+P(1)+P(2)+P(3)\).
  • 2
    \(P(0)=e^{-2}\approx 0.1353\)
    \(P(1)=2e^{-2}\approx 0.2707\)
    \(P(2)=\frac{4e^{-2}}{2}\approx 0.2707\)
    \(P(3)=\frac{8e^{-2}}{6}\approx 0.1804\)
  • 3
    \(P(X\leq 3)=0.1353+0.2707+0.2707+0.1804=0.8571\)
Answer: P(X ≤ 3) ≈ 0.8571
Practice Problems
20 Exam-Style Questions
Problem 1 Direct PMF

A call centre receives an average of 3 calls per minute. The number of calls follows a Poisson distribution.

Find \(P(X = 2)\). Give your answer to 4 decimal places.

Problem 2 PMF: Zero Events

A Geiger counter records an average of 5 alpha particles per second from a radioactive source.

Find the probability that no alpha particles are recorded in a given second. Give your answer to 4 decimal places.

Problem 3 Rate Scaling

Emails arrive at a server at an average rate of 12 per hour.

Find the probability that exactly 1 email arrives in a 10-minute period. Give your answer to 4 decimal places.

Problem 4 Cumulative: At Most

A road intersection has an average of 2 accidents per month. Assume a Poisson model.

Find \(P(X \leq 2)\) for a randomly selected month. Give your answer to 4 decimal places.

Problem 5 Complement Rule

Typos in a manuscript occur at an average rate of 1.5 per page.

Find the probability that a randomly chosen page contains at least one typo. Give your answer to 4 decimal places.

Problem 6 P(X > k)

A bank ATM machine has breakdowns occurring at an average rate of 0.8 per week.

Find the probability that the ATM has more than 2 breakdowns in a given week. Give your answer to 4 decimal places.

Problem 7 Mean & Variance

The number of goals scored per match in a football league follows \(X \sim \text{Po}(\lambda)\). The variance of goals scored is 2.7.

State the value of \(\lambda\) and find the standard deviation of \(X\). Give the standard deviation to 4 decimal places.

Problem 8 Rate Scaling (Area)

Flaws in a sheet of glass occur at an average rate of 3 per square metre. A sheet of glass has an area of 0.5 m².

Find the probability that the sheet contains exactly 2 flaws. Give your answer to 4 decimal places.

Problem 9 Sum of Poissons

Customers arrive at checkout A at an average rate of 2 per minute and at checkout B at an average rate of 3 per minute, independently of each other.

Find the probability that a total of exactly 4 customers arrive at both checkouts combined in one minute. Give your answer to 4 decimal places.

Problem 10 Binomial Approximation

In a large city, the probability that any one person is left-handed is 0.02. A random sample of 200 people is selected.

Using a Poisson approximation, find the probability that exactly 5 people in the sample are left-handed. Give your answer to 4 decimal places.

Problem 11 Conditional Probability

Website errors occur at a rate of 2 per hour. Given that at least 1 error occurs in an hour, find the probability that exactly 2 errors occur.

Find \(P(X=2 \mid X \geq 1)\). Give your answer to 4 decimal places.

Problem 12 Finding Lambda

The number of misprints per page of a book follows \(X \sim \text{Po}(\lambda)\). It is known that \(P(X = 0) = 0.4066\).

Find the value of \(\lambda\). Give your answer to 4 decimal places.

Problem 13 Multi-Interval

Buses arrive at a stop at an average rate of 4 per hour.

Find the probability that at least 3 buses arrive in a 30-minute period. Give your answer to 4 decimal places.

Problem 14 Expected Value Application

A factory produces bolts. On average, 0.5% of bolts are defective. A sample of 1000 bolts is examined. Let \(X\) be the number of defective bolts.

Using a Poisson approximation, find \(E[X]\) and \(P(X \geq 3)\). Give \(P(X \geq 3)\) to 4 decimal places.

Problem 15 P(a ≤ X ≤ b)

The number of earthquakes of magnitude 4.0 or greater in a region follows \(\text{Po}(3)\) per year.

Find the probability that between 2 and 4 earthquakes (inclusive) occur in a given year. Give your answer to 4 decimal places.

Problem 16 Mode of Poisson

The number of daily power outages in a district follows \(X \sim \text{Po}(4)\).

Find the mode(s) of \(X\). Then calculate \(P(X = \text{mode})\) for the smaller mode if two modes exist. Give your probability to 4 decimal places.

Problem 17 Independent Poissons

Machine A breaks down at an average rate of 1 per week. Machine B breaks down at an average rate of 2 per week. Breakdowns are independent.

Find the probability that the total number of breakdowns in one week is exactly 3. Give your answer to 4 decimal places.

Problem 18 Inequality / At Least

A random variable \(X \sim \text{Po}(6)\).

Find \(P(X \geq 5)\). Give your answer to 4 decimal places.

Problem 19 Estimating Lambda from Data

A biologist counts bacteria colonies on 100 Petri dishes. The data is: 0 colonies (30 dishes), 1 colony (36 dishes), 2 colonies (22 dishes), 3 colonies (9 dishes), 4+ colonies (3 dishes).

Estimate \(\hat{\lambda}\) using the sample mean. Give your answer to 4 decimal places.

Problem 20 Multi-Step Applied Problem

A customer service hotline receives calls at an average rate of 5 per 15 minutes. Assuming a Poisson model:

(a) Find the probability that exactly 3 calls are received in a 6-minute period.
(b) Find the probability of receiving fewer than 2 calls in a 6-minute period.

Enter the answer to part (a) to 4 decimal places.

Answer Key & Full Solutions
Detailed Solutions
Problem 1 Direct PMF  |  \(\lambda=3\)
Answer: 0.2240
\(X\sim\text{Po}(3)\).   \(P(X=2)=\dfrac{e^{-3}\cdot 3^2}{2!}=\dfrac{0.049787\times 9}{2}=\dfrac{0.44809}{2}\approx 0.2240\)
Problem 2 PMF: Zero Events  |  \(\lambda=5\)
Answer: 0.0067
\(P(X=0)=\dfrac{e^{-5}\cdot 5^0}{0!}=e^{-5}\approx 0.006738 \approx 0.0067\)
Problem 3 Rate Scaling  |  10 min interval
Answer: 0.2707
Rate: 12 per hour. 10 min = 10/60 hour. New \(\lambda=12\times\frac{10}{60}=2\).
\(P(X=1)=\dfrac{e^{-2}\cdot 2^1}{1!}=2e^{-2}\approx 2\times 0.13534=0.2707\)
Problem 4 Cumulative \(P(X\leq 2)\)  |  \(\lambda=2\)
Answer: 0.6767
\(P(0)=e^{-2}\approx0.1353\);  \(P(1)=2e^{-2}\approx0.2707\);  \(P(2)=\frac{4e^{-2}}{2}\approx0.2707\)
\(P(X\leq 2)=0.1353+0.2707+0.2707=0.6767\)
Problem 5 Complement  |  \(\lambda=1.5\)
Answer: 0.7769
\(P(X\geq 1)=1-P(X=0)=1-e^{-1.5}=1-0.22313\approx 0.7769\)
Problem 6 \(P(X>2)\)  |  \(\lambda=0.8\)
Answer: 0.0474
\(P(X>2)=1-P(X\leq 2)\)
\(P(0)=e^{-0.8}\approx0.4493\);  \(P(1)=0.8e^{-0.8}\approx0.3595\);  \(P(2)=\frac{0.64e^{-0.8}}{2}\approx0.1438\)
\(P(X\leq2)\approx0.9526\);  \(P(X>2)\approx1-0.9526=0.0474\)
Problem 7 Mean & Variance  |  \(\lambda=2.7\)
Answer: \(\lambda=2.7\);   \(\sigma\approx 1.6432\)
For Poisson, \(\text{Var}(X)=\lambda\), so \(\lambda=2.7\).
\(\sigma=\sqrt{\lambda}=\sqrt{2.7}\approx 1.6432\)
Problem 8 Area Scaling  |  \(\lambda'=1.5\)
Answer: 0.2510
Rate: 3 per m². Area: 0.5 m². Scaled rate: \(\lambda'=3\times 0.5=1.5\).
\(P(X=2)=\dfrac{e^{-1.5}\cdot(1.5)^2}{2!}=\dfrac{0.22313\times2.25}{2}=\dfrac{0.50204}{2}\approx 0.2510\)
Problem 9 Sum of Poissons  |  \(\lambda=5\)
Answer: 0.1755
\(A\sim\text{Po}(2)\), \(B\sim\text{Po}(3)\) independent. \(T=A+B\sim\text{Po}(5)\).
\(P(T=4)=\dfrac{e^{-5}\cdot 5^4}{4!}=\dfrac{0.006738\times625}{24}=\dfrac{4.2112}{24}\approx 0.1755\)
Problem 10 Binomial Approx.  |  \(\lambda=4\)
Answer: 0.1563
\(n=200,\ p=0.02\). Since \(n\geq50\) and \(np=4\leq5\), use \(X\sim\text{Po}(4)\).
\(P(X=5)=\dfrac{e^{-4}\cdot4^5}{5!}=\dfrac{0.01832\times1024}{120}=\dfrac{18.761}{120}\approx 0.1563\)
Problem 11 Conditional Probability  |  \(\lambda=2\)
Answer: 0.2998
\(P(X=2\mid X\geq1)=\dfrac{P(X=2)}{P(X\geq1)}=\dfrac{P(X=2)}{1-P(X=0)}\)
\(P(X=2)=\dfrac{e^{-2}\cdot4}{2}=2e^{-2}\approx0.2707\)
\(P(X\geq1)=1-e^{-2}\approx1-0.1353=0.8647\)
\(P(X=2\mid X\geq1)=\dfrac{0.2707}{0.8647}\approx 0.3130\)
Problem 12 Finding \(\lambda\)
Answer: \(\lambda \approx 0.8994\)
\(P(X=0)=e^{-\lambda}=0.4066\)
\(-\lambda=\ln(0.4066)\approx-0.8994\)
\(\lambda\approx 0.8994\)
Problem 13 Multi-Interval  |  \(\lambda'=2\)
Answer: 0.3233
Rate: 4/hr. 30 min = 0.5 hr. \(\lambda'=4\times0.5=2\).
\(P(X\geq3)=1-P(X\leq2)=1-[e^{-2}+2e^{-2}+2e^{-2}]=1-5e^{-2}\)
\(=1-5\times0.13534=1-0.6767=0.3233\)
Problem 14 Poisson Approx.  |  \(\lambda=5\)
Answer: \(E[X]=5\);   \(P(X\geq3)\approx 0.8753\)
\(n=1000,\ p=0.005\). \(\lambda=np=5\). So \(X\sim\text{Po}(5)\).
\(P(X\geq3)=1-P(X\leq2)\)
\(P(0)=e^{-5}\approx0.0067\); \(P(1)=5e^{-5}\approx0.0337\); \(P(2)=12.5e^{-5}\approx0.0842\)
\(P(X\leq2)\approx0.1247\);  \(P(X\geq3)\approx 0.8753\)
Problem 15 \(P(2\leq X\leq4)\)  |  \(\lambda=3\)
Answer: 0.6161
\(P(2)+P(3)+P(4)\) where \(\lambda=3\):
\(P(2)=\frac{9e^{-3}}{2}\approx0.2240\);  \(P(3)=\frac{27e^{-3}}{6}\approx0.2240\);  \(P(4)=\frac{81e^{-3}}{24}\approx0.1680\)
Total \(\approx0.2240+0.2240+0.1680=0.6161\) (Note: slight rounding — exact sum \(\approx0.6161\))
Problem 16 Mode of Poisson  |  \(\lambda=4\)
Answer: Mode(s) = 3 and 4;   \(P(X=3)\approx 0.1954\)
When \(\lambda\) is a positive integer, Poisson has two modes: \(\lambda-1\) and \(\lambda\). Here modes are 3 and 4.
\(P(X=3)=\dfrac{e^{-4}\cdot4^3}{3!}=\dfrac{0.01832\times64}{6}\approx 0.1954\)
Verify: \(P(X=4)=\dfrac{e^{-4}\cdot4^4}{4!}=\dfrac{0.01832\times256}{24}\approx 0.1954\) ✓
Problem 17 Independent Poissons  |  \(\lambda=3\)
Answer: 0.2240
\(A\sim\text{Po}(1)\), \(B\sim\text{Po}(2)\). \(T=A+B\sim\text{Po}(3)\).
\(P(T=3)=\dfrac{e^{-3}\cdot3^3}{3!}=\dfrac{0.04979\times27}{6}=\dfrac{1.3443}{6}\approx 0.2240\)
Problem 18 \(P(X\geq5)\)  |  \(\lambda=6\)
Answer: 0.7149
\(P(X\geq5)=1-P(X\leq4)\)
\(P(0)=e^{-6}\approx0.002479\)
\(P(1)=6e^{-6}\approx0.014873\)
\(P(2)=18e^{-6}\approx0.044618\)
\(P(3)=36e^{-6}\approx0.089235\)
\(P(4)=54e^{-6}\approx0.133853\)
\(P(X\leq4)\approx0.28513\);  \(P(X\geq5)\approx1-0.28513=0.7149\)
Problem 19 MLE of \(\lambda\)
Answer: \(\hat{\lambda}\approx 1.1900\)
Assume 4+ colonies = exactly 4 (conservative estimate for the 3 dishes).
Total colonies \(= 0\times30 + 1\times36 + 2\times22 + 3\times9 + 4\times3 = 0+36+44+27+12=119\)
\(\hat{\lambda}=\dfrac{119}{100}=1.19\)
Problem 20 Multi-Step  |  \(\lambda'=2\)
Part (a): 0.1804  |  Part (b): 0.4060
Rate: 5 per 15 min = 1/3 per min. Over 6 min: \(\lambda'=6\times\frac{1}{3}=2\).
(a) \(P(X=3)=\dfrac{e^{-2}\cdot8}{6}=\dfrac{0.13534\times8}{6}=\dfrac{1.0827}{6}\approx 0.1804\)
(b) \(P(X<2)=P(0)+P(1)=e^{-2}+2e^{-2}=3e^{-2}\approx3\times0.13534=0.4060\)
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