📞 Poisson Distribution
When events happen randomly over time or space at a known average rate — component failures per month, flaws per metre of cable — the Poisson distribution predicts how many will occur in a given interval.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
✅Conditions & FormulaP(X=r) = e⁻ᵐmʳ/r!
🔗Mean, Variance & Binomial LinkWhen to approximate
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Recognise when events fit the Poisson model; calculate Poisson probabilities using the mean rate m; and understand when the Poisson distribution can be used to approximate a binomial distribution.
💡 Poisson needs only one parameter — the mean rate m — unlike the binomial's n and p. It's the natural model whenever you're counting rare, random, independent events over a fixed interval.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
✅
Poisson Conditions & Formula
A distribution for counting rare, random events — needing only the average rate.
🔧Poisson Probability CalculatorEnter m and r
Enter m and r, then press Calculate to see e⁻ᵐ, mʳ, r! and the final probability…
📖 Key Vocabulary
m (or λ) — the mean (average) number of events per interval; the single parameter of the distribution.
e — Euler's number ≈ 2.71828, appearing because Poisson is derived as a limiting case of the binomial.
Interval — the fixed unit of time, length or area over which events are counted (e.g. per hour, per metre).
Rare event — Poisson suits events with low individual probability but many opportunities to occur.
👁️ Visual aid: unlike the binomial, r has no upper limit — the distribution has a long right tail even though most probability sits near the mean.
✏️Worked Example 1 — Direct probability calculationType: Direct calculation
Question A component fails on average 2 times per year (m=2). Find P(X=1) in a given year.
Substitute
P(X=1) = e⁻² × 2¹ / 1!Compute
= 0.1353 × 2 / 1Answer P(X=1) ≈ 0.2707 (4 d.p.)
0️⃣Worked Example 2 — The special case P(X=0)Type: Zero events
Question For m=1.5, find the probability of zero events.
Substitute
P(X=0) = e⁻¹·⁵ × 1.5⁰ / 0!Simplify
1.5⁰=1, 0!=1, so P(X=0) = e⁻¹·⁵Answer P(X=0) ≈ 0.2231 (4 d.p.)
➕Worked Example 3 — Cumulative probabilityType: Cumulative
Question For m=2, find P(X≤1).
P(X=0)
e⁻² = 0.1353P(X=1)
e⁻² × 2 = 0.2707Answer P(X≤1) = 0.1353+0.2707 = 0.4060 (4 d.p.)
⚙️Worked Example 4 — Engineering context (call centre faults)Type: Engineering
Question A production line averages 1.8 minor stoppages per shift. Find the probability of exactly 3 stoppages in a shift.
Set up
m=1.8, r=3Substitute
P(X=3) = e⁻¹·⁸ × 1.8³ / 3! = 0.1653 × 5.832 / 6Answer P(X=3) ≈ 0.1607 (4 d.p.) — useful for staffing decisions around expected stoppage-handling workload.
🧠 Common slip: forgetting r! in the denominator, especially for r=0 (0!=1, not 0) — always double check factorial values before dividing.
🔗
Mean, Variance & the Binomial Link
Poisson's mean and variance are both m — and it can stand in for the binomial when n is large and p is small.
🔧Poisson Approximation to BinomialEnter n, p, r — compares both methods
Enter n, p and r, then press Compare to see the exact binomial probability side by side with the Poisson approximation…
📖 Key Vocabulary
Poisson approximation — using Poisson (with m=np) to estimate a binomial probability when direct calculation is impractical.
Rule of thumb — approximation works well when n is large (≥50) and p is small (np<5).
Equidispersion — the unusual property that Poisson's mean equals its variance.
Sum of Poisson variables — if two independent Poisson processes have means m₁, m₂, their sum is Poisson with mean m₁+m₂.
👁️ Visual aid: as n grows and p shrinks while np stays fixed, the binomial's shape converges almost exactly onto the Poisson distribution.
✏️Worked Example 1 — Mean and varianceType: Direct calculation
Question For a Poisson distribution with m=4, state the mean and variance.
Apply the rule Mean = m, Variance = m
Answer Mean = 4, Variance = 4, SD = √4 = 2 — no separate variance formula is needed.
🔗Worked Example 2 — Checking approximation validityType: Validity check
Question A binomial has n=300, p=0.01. Can Poisson be used to approximate it? If so, find m.
Check n=300≥50 ✓, np=300×0.01=3<5 ✓
Answer Yes, approximation is valid; use m = np = 3.
📊Worked Example 3 — Comparing exact binomial to Poisson estimateType: Comparison
Question For n=100, p=0.02, estimate P(X=2) using Poisson (m=np=2), and note this closely matches the true binomial value.
Poisson estimate
P(X=2) = e⁻² × 2² / 2! = 0.1353 × 4/2Answer P(X=2) ≈ 0.2707 — the true binomial value is 0.2734, a difference of under 0.003, showing how good the approximation is here.
⚙️Worked Example 4 — Engineering context (rare defects in large batches)Type: Engineering
Question A factory produces 1000 PCBs; each has a 0.3% chance of a solder defect. Estimate the probability of exactly 4 defective boards using Poisson.
Check validity n=1000≥50 ✓, np=1000×0.003=3<5 ✓ — Poisson is appropriate
Substitute
m=3, P(X=4) = e⁻³ × 3⁴/4! = 0.0498 × 81/24Answer P(X=4) ≈ 0.1680 (4 d.p.) — computing this directly from the binomial formula with n=1000 would be far more cumbersome.
🧠 Common slip: using the Poisson approximation when p is not small (e.g. p=0.3) — even with large n, the shapes diverge badly unless np stays modest (below about 5).
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔌 Equipment Failures per Month
Failures occurring randomly and independently over time are a classic Poisson scenario for maintenance planning.
P(X=r) = e⁻ᵐ mʳ / r!
📟 Try it — Failure Probability Calculator
Enter the mean rate and r, then press Calculate…
🧵 Fabric/Cable Flaws per Length
Manufacturing flaws distributed randomly along a length of material (cable, wire, textile) follow a Poisson process per metre.
m scales with length: m = rate per metre × length
📟 Try it — Flaw Probability Calculator
Enter the rate, length and r, then press Calculate…
📞 Call Centre / Helpdesk Arrivals
Service requests arriving randomly at a known average rate are modelled directly with Poisson for staffing decisions.
P(X=r) = e⁻ᵐ mʳ / r!
📟 Try it — Arrivals Calculator
Enter the mean call rate and r, then press Calculate…
🔬 Rare Defect Estimation (Poisson Approximation)
For very large batches with a tiny per-unit defect rate, Poisson gives a fast, accurate estimate without a full binomial calculation.
m = n × p (valid when n≥50, np<5)
📟 Try it — Large-Batch Defect Estimator
Enter n, p and r, then press Calculate…
🎮
Quick-Fire Poisson Game
12 questions on conditions, probability and the binomial link. How many can you get?
⚡ Poisson Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
For m=3, state the variance.
Variance = m = 3
Question 2 🥉2 marks
For m=1, find P(X=0).
e⁻¹×1⁰/0! = e⁻¹ = 0.3679
Question 3 🥉2 marks
State the rule of thumb for using Poisson to approximate a binomial.
n large (≥50) and np small (<5)
Question 4 🥉2 marks
For n=400, p=0.005, find m for the Poisson approximation.
m = np = 400×0.005 = 2
Question 5 🥈3 marks
For m=2.5, find P(X=2).
e⁻²·⁵×2.5²/2! = 0.0821×6.25/2
≈0.2565
≈0.2565
Question 6 🥈4 marks
For m=1.2, find P(X≤1).
P(X=0)=e⁻¹·²=0.3012
P(X=1)=e⁻¹·²×1.2=0.3614
Sum=0.3012+0.3614
=0.6626
P(X=1)=e⁻¹·²×1.2=0.3614
Sum=0.3012+0.3614
=0.6626
Question 7 🥈3 marks
A workshop averages 0.8 accidents per month. Find P(X=0) for a given month.
e⁻⁰·⁸ = 0.4493
Question 8 🥈4 marks
For m=3.5, find P(X=4).
e⁻³·⁵×3.5⁴/4! = 0.0302×150.0625/24
≈0.1888
≈0.1888
Question 9 🥇4 marks
A call centre receives 4.5 calls per minute on average. Find the probability of exactly 6 calls in a minute.
m=4.5,r=6
e⁻⁴·⁵×4.5⁶/6! = 0.01111×8303.77/720
≈0.1281
e⁻⁴·⁵×4.5⁶/6! = 0.01111×8303.77/720
≈0.1281
Question 10 🥇5 marks
A cable manufacturer finds 0.4 flaws per 10m section on average. Find the probability of at least 1 flaw in a 10m section.
P(X=0)=e⁻⁰·⁴=0.6703
P(X≥1)=1−0.6703
=0.3297
P(X≥1)=1−0.6703
=0.3297
Question 11 🥇5 marks
A factory has n=800 units with a defect rate of 0.25%. Use Poisson to estimate P(exactly 3 defective).
m=800×0.0025=2
P(X=3)=e⁻²×2³/3! = 0.1353×8/6
≈0.1804
P(X=3)=e⁻²×2³/3! = 0.1353×8/6
≈0.1804
Question 12 🥇5 marks
Two independent Poisson processes have means 1.5 and 2.5 faults per day for two separate machines. Find the probability of exactly 2 total faults across both machines combined.
Combined mean = 1.5+2.5 = 4 (sum of independent Poisson variables)
P(X=2) = e⁻⁴×4²/2! = 0.01832×16/2
≈0.1465
P(X=2) = e⁻⁴×4²/2! = 0.01832×16/2
≈0.1465