🎲 Binomial Distribution
When every trial has just two outcomes — pass/fail, defective/good — and the probability stays fixed, the binomial distribution predicts exactly how many "successes" to expect out of n attempts.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
✅Conditions & FormulaP(X=r) = ⁿCᵣpʳq⁽ⁿ⁻ʳ⁾
📊Mean, Variancenp and npq
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Recognise when a scenario satisfies the binomial conditions; calculate individual and cumulative binomial probabilities; and find the mean and variance of a binomial distribution using np and npq.
💡 Four conditions must all hold: fixed number of trials n, only two outcomes each time, constant probability p, and independent trials. If any one fails, the binomial model doesn't apply.
🧭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
✅
Binomial Conditions & Formula
Four conditions must hold before this model applies — then one formula answers every probability question.
🔧Binomial Probability CalculatorEnter n, p and r
Enter n, p and r, then press Calculate to see ⁿCᵣ, pʳ, q⁽ⁿ⁻ʳ⁾ and the final probability…
📖 Key Vocabulary
Trial — a single repetition of the experiment (e.g. testing one component).
Success/failure — the two possible outcomes of each trial (labels are arbitrary — "success" need not be good news).
p and q — p is the probability of success, q = 1−p is the probability of failure.
ⁿCᵣ (binomial coefficient) — counts how many different orderings give exactly r successes.
👁️ Visual aid: the tallest bars cluster around the expected value np — here 10×0.3=3 successes is the single most likely outcome.
✏️Worked Example 1 — Checking the conditionsType: Recognition
Question A batch of 20 components is tested; each has a 5% chance of being defective, independently. Confirm this is binomial and state n, p, q.
Check Fixed n=20 ✓, two outcomes (defective/good) ✓, constant p=0.05 ✓, independent ✓
Answer Binomial applies: n=20, p=0.05, q=0.95
🔢Worked Example 2 — Calculating a single probabilityType: Direct calculation
Question For n=6, p=0.4, find P(X=2).
Coefficient
⁶C₂ = 6!/(2!4!) = 15Substitute
P(X=2) = 15 × 0.4² × 0.6⁴ = 15 × 0.16 × 0.1296Answer P(X=2) ≈ 0.3110 (4 d.p.)
➕Worked Example 3 — Cumulative probability ("at least")Type: Cumulative
Question For n=5, p=0.3, find P(X≥1).
Use the complement
P(X≥1) = 1 − P(X=0)Find P(X=0)
⁵C₀(0.3)⁰(0.7)⁵ = 1×1×0.16807Answer P(X≥1) = 1 − 0.16807 = 0.8319 (4 d.p.)
⚙️Worked Example 4 — Engineering context (batch acceptance)Type: Engineering
Question A sample of 8 welds is inspected; each has a 10% chance of failing. Find the probability that exactly 2 fail.
Set up
n=8, p=0.1, r=2. ⁸C₂ = 28Substitute
P(X=2) = 28 × 0.1² × 0.9⁶ = 28 × 0.01 × 0.531441Answer P(X=2) ≈ 0.1488 (4 d.p.) — roughly a 15% chance of exactly 2 weld failures in this sample.
🧠 Common slip: using the binomial model when p changes between trials (e.g. sampling without replacement from a small population) — that violates the "constant p" condition and the hypergeometric distribution should be used instead.
📊
Mean, Variance & Applications
Two simple formulas summarise the whole distribution's centre and spread.
🔧Mean & Variance CalculatorEnter n and p
Enter n and p, then press Calculate for the mean, variance and standard deviation…
📖 Key Vocabulary
Expected value — another name for the mean, np; the long-run average number of successes.
Skewness — asymmetry of the distribution; the binomial is skewed unless p=0.5.
Discrete distribution — X can only take whole-number values 0,1,2,…,n.
Normal approximation — for large n, the binomial shape approaches a normal curve (useful for estimation).
👁️ Visual aid: a small p piles probability near r=0, while p=0.5 gives a perfectly symmetric, bell-like shape.
✏️Worked Example 1 — Mean and varianceType: Direct calculation
Question For n=50, p=0.2, find the mean, variance and SD.
Mean
np = 50×0.2 = 10Variance
npq = 50×0.2×0.8 = 8Answer Mean=10, Variance=8, SD=√8≈2.83
📈Worked Example 2 — Interpreting the mean in contextType: Interpretation
Question A machine produces items with a 3% defect rate. In a batch of 200, how many defects are expected on average?
Apply
np = 200 × 0.03Answer Expected defects = 6 per batch of 200, on average.
📐Worked Example 3 — Comparing spread for different pType: Comparison
Question For n=100, compare the SD when p=0.1 versus p=0.5.
p=0.1
√(100×0.1×0.9) = √9 = 3p=0.5
√(100×0.5×0.5) = √25 = 5Answer Variability is greatest when p=0.5 (maximum uncertainty) and smaller when p is close to 0 or 1 (more predictable outcomes).
⚙️Worked Example 4 — Engineering context (production planning)Type: Engineering
Question A factory runs 500 units through a test with a 4% failure rate. Find the expected number of failures and the SD, and use them to estimate a realistic range.
Mean
np = 500×0.04 = 20SD
√(500×0.04×0.96) = √19.2 ≈ 4.38Answer Expect around 20 failures, typically within roughly ±2SD (≈11 to 29) — useful for planning rework capacity without over- or under-resourcing.
🧠 Common slip: confusing variance (npq) with standard deviation (√npq) — always check which one a question asks for, since they're numerically very different.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔩 Batch Acceptance Sampling
Quality inspectors sample n items from a batch and accept it only if defects found are below a threshold — a direct binomial application.
P(X=r) = ⁿCᵣ pʳ q⁽ⁿ⁻ʳ⁾
📟 Try it — Acceptance Sampling Calculator
Enter n, p and r, then press Calculate…
🔌 Component Reliability — Number Surviving
If each component has a known survival probability, the binomial model predicts how many out of a batch will still be working.
Expected survivors = n × p(survival)
📟 Try it — Survivor Estimator
Enter batch size and survival probability, then press Calculate…
📡 Signal Transmission — Bit Error Probability
Each transmitted bit has a small independent chance of error; binomial probability gives the chance of exactly r errors in n bits.
P(exactly r errors) = ⁿCᵣ pʳ q⁽ⁿ⁻ʳ⁾
📟 Try it — Bit Error Calculator
Enter n, p and r, then press Calculate…
🏭 Maintenance Planning — Expected Failures
Using np and npq lets planners size a maintenance team based on both the expected workload and its realistic variability.
Mean = np, SD = √(npq)
📟 Try it — Maintenance Planner
Enter units in service and failure probability, then press Calculate…
🎮
Quick-Fire Binomial Game
12 questions on conditions, probability and mean/variance. How many can you get?
⚡ Binomial 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 n=4, p=0.5, find ⁴C₂.
4!/(2!2!) = 24/4 = 6
Question 2 🥉2 marks
For n=10, p=0.3, find the mean.
np = 10×0.3 = 3
Question 3 🥉2 marks
For n=8, p=0.25, find the variance.
npq = 8×0.25×0.75 = 1.5
Question 4 🥉2 marks
State the four conditions required for a binomial distribution.
Fixed number of trials; only two outcomes; constant probability p; independent trials
Question 5 🥈3 marks
For n=5, p=0.4, find P(X=3).
⁵C₃=10
P=10×0.4³×0.6²=10×0.064×0.36
=0.2304
P=10×0.4³×0.6²=10×0.064×0.36
=0.2304
Question 6 🥈4 marks
For n=6, p=0.5, find P(X=0).
⁶C₀=1
P=1×1×0.5⁶=1/64
=0.0156
P=1×1×0.5⁶=1/64
=0.0156
Question 7 🥈3 marks
For n=12, p=0.15, find the mean and SD.
Mean=12×0.15=1.8
Var=12×0.15×0.85=1.53
SD=√1.53≈1.237
Var=12×0.15×0.85=1.53
SD=√1.53≈1.237
Question 8 🥈4 marks
For n=7, p=0.2, find P(X≥1) using the complement.
P(X=0)=0.8⁷=0.2097
P(X≥1)=1−0.2097
=0.7903
P(X≥1)=1−0.2097
=0.7903
Question 9 🥇4 marks
A batch of 25 items has a 6% defect rate. Find the probability that exactly 3 are defective.
n=25,p=0.06,r=3
²⁵C₃=2300
P=2300×0.06³×0.94²²
=2300×0.000216×0.2564
≈0.1273
²⁵C₃=2300
P=2300×0.06³×0.94²²
=2300×0.000216×0.2564
≈0.1273
Question 10 🥇5 marks
A machine has a 2% fault rate per unit. In a run of 150 units, find the expected number of faults and the SD.
Mean=150×0.02=3
Var=150×0.02×0.98=2.94
SD=√2.94≈1.715
Var=150×0.02×0.98=2.94
SD=√2.94≈1.715
Question 11 🥇5 marks
For n=10, p=0.35, find P(X≤1) (i.e. P(X=0)+P(X=1)).
P(X=0)=0.65¹⁰=0.01346
P(X=1)=10×0.35×0.65⁹=10×0.35×0.02071=0.07249
Sum=0.01346+0.07249
≈0.0860
P(X=1)=10×0.35×0.65⁹=10×0.35×0.02071=0.07249
Sum=0.01346+0.07249
≈0.0860
Question 12 🥇5 marks
A component has a 92% reliability (probability of working correctly). In a system using 6 of these components independently, find the probability that at least 5 work.
n=6,p=0.92
P(X=6)=0.92⁶=0.6064
P(X=5)=⁶C₅×0.92⁵×0.08=6×0.6592×0.08=0.3164
P(X≥5)=0.6064+0.3164
≈0.9228
P(X=6)=0.92⁶=0.6064
P(X=5)=⁶C₅×0.92⁵×0.08=6×0.6592×0.08=0.3164
P(X≥5)=0.6064+0.3164
≈0.9228