A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
A solar-panel factory tests 20 cells from a production line. Each cell independently passes calibration with probability 0.85. The line should average 17 passing cells per group—but how often will exactly 17 pass? How unusual would fewer than 10 passes be?
An average alone cannot answer those questions. We need a model for the entire pattern of possible success counts: the binomial distribution.
You already have both ingredients. L-PR-3 supplied combinations for counting arrangements, and L-PR-4 supplied probability distributions, cumulative events, means, and standard deviations. This lesson fuses them into one reusable model.
After this lesson, you will be able to:
Check BINS conditions and decide whether a situation is binomial.
Compute and interpret exact and cumulative binomial probabilities.
Calculate and interpret ,, and distribution shape from and .
Diagnose a non-binomial setting or flawed binomial analysis and correct it.
The binomial distribution appears in quality control, genetics, polling, and any other setting with a fixed number of independent trials and a constant success probability.
Section 2: Prerequisites
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
What you need coming into this lesson:
L-PR-4 — PMF validity: A probability mass function satisfies and . The binomial formula creates a special PMF, so these rules still apply.
Retrieval Checkpoint — warm up the machinery you’ll need.
Problem 1.
Success Factor — bridge from old rules to the new model.
If the combination was difficult, revisit L-PR-3, Section 3. If cumulative-event language or expected value feels uncertain, revisit L-PR-4, Section 3.
L-PR-4 built each PMF from first principles. L-PR-5 introduces a named family whose formula replaces a list of all success/failure sequences—but only after the four BINS conditions are checked. That check is where Core Concept C1 begins.
Section 3: Core Concepts
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Navigation — six concepts in this section:
C1 — BINS Conditions: The four requirements that define a binomial setting.
C5 — Shape of the Distribution: Skewness as a function of ; effect of .
C6 — Recognizing Non-Binomial Settings: When to reject the binomial model.
C0 — Start With Counts: What Happens Over Many Trials
Before the formula, think in frequencies.
Return to the solar-cell line from the introduction. Each of 20 cells independently passes calibration with probability 0.85. How many passes do we expect?
The naive answer is . But “17 passing cells” is just the average over many repetitions of this process. In any one group of 20 cells, you might get 14 passes, or 18, or 16. The question is: what’s the complete picture of all possible outcomes?
Think of each inspection as an independent coin flip (heads = pass) with . Twenty flips produce a count of passes. That count changes from group to group; its probability pattern is the binomial distribution.
The exact binomial model predicts:
, so about 24.3% of groups have exactly 17 passing cells
, so about 99.4% have between 13 and 20 passes
, fewer than 4 groups in 100,000
The binomial distribution gives us the exact probability for every possible count. This is more useful than just knowing the average — it tells quality teams what to expect in practice, not just in theory.
C1 — BINS Conditions
Before applying the binomial formula, verify that your situation satisfies all four conditions. We remember them with the mnemonic BINS:
BINS Conditions for a Binomial Setting
A random variable follows a binomial distribution if and only if:
B — Binary outcomes: Each trial results in exactly one of two outcomes, called “success” and “failure.” (The labels are arbitrary — a defective item can be the “success” if that’s what you’re counting.)
I — Independent trials: The outcome of any one trial does not affect the probability of success on any other trial.
N — Fixed number of trials: The total number of trials, , is determined in advance and does not change.
S — Constant success probability: The probability of success, , is the same on every trial.
We write to mean ” is a binomial random variable with parameters and .”
Mini-example — checking BINS for a free-throw scenario.
A basketball player makes each free throw independently with probability 0.75. She attempts 8 free throws in a game. Let = number made.
B: Each free throw is either made (success) or missed (failure). ✓
I: Each attempt is independent — she doesn’t “get tired” or “get hot” in this idealized model. ✓
N: is fixed in advance. ✓
S: on every attempt. ✓
All four hold: .
Every condition must hold. If even one fails, you cannot use the binomial formula. A common error is to check only Binary and Independent while forgetting to verify that is fixed and is constant. For instance, if a student keeps attempting free throws until she makes 3 (no fixed ), the binomial model does not apply — that is a negative binomial setting. Waiting until the first success is the special geometric case.
C2 — Binomial Probability Formula
Once BINS conditions are verified, the probability that exactly successes occur in trials is:
Binomial Probability Formula
where is the number of ways to arrange successes among trials.
We write for the failure probability, so the formula becomes . Always establish explicitly — do not substitute a value for without first writing .
Why is the combination coefficient there? Consider trials and successes. The possible sequences of successes (S) and failures (F) are: SSF, SFS, FSS. Each sequence has probability . There are such sequences, so the total probability is .
Generalizing: there are sequences with exactly successes, each with probability . Multiplying gives the formula.
p = 0.40, p²(1−p) ≈ 0.0960, P(X = 2) ≈ 0.2880
Before dragging: will the 3 change when p changes?
Figure 1: All 8 possible sequences for n = 3 trials. Highlighted rows each contain exactly
k = 2 successes (S) and 1 failure (F). The bracket shows that C(3,2) = 3 such sequences exist
— this is the combination coefficient in the binomial formula. Drag p to see how the probability
of each sequence changes while the count of 3 stays fixed.
Mini-example — free-throw scenario continued.
Find for .
There is approximately a 31.1% chance that she makes exactly 6 of 8 free throws.
The most common error in the entire lesson: writing without . This formula gives the probability of one specific sequence with successes (e.g., SSSSSFF), not the probability that exactly successes occur in any order. Always include the combination coefficient.
C3 — Cumulative Binomial Probabilities
A single application of the binomial formula gives a point probability — the chance of exactly successes. Real problems often ask for ranges: “at most 3 broken items,” “at least 5 correct answers.” These require summing terms.
Cumulative Binomial Probability
For :
Phrasing
Formal form
How to compute
“exactly ”
One term of the formula
“at most ”
“fewer than ”
“at least ”
“more than ”
The complement identity is particularly useful when summing from to would require many terms.
P(X ≤ 3) ≈ 0.5941 for X ∼ B(8, 0.40)
Figure 2: X ∼ B(8, 0.40) PMF. Blue bars are included in the selected probability; grey bars are
excluded. The green border marks the boundary value k — solid when k is included,
dashed when excluded. Toggle between "at most k" and "fewer than k" at the
same k to see exactly which bar shifts.
Mini-example — cumulative boundaries.
. Compare “at least 3” vs. “more than 3."
"At least 3” = . Boundary: 3 is included.
”More than 3” = . Boundary: 3 is excluded.
These differ by exactly one term: . The ±1 boundary shift is where most errors occur.
“At least 1” does NOT equal “exactly 1.” . Students who read “at least 1” as “exactly 1” will compute a point probability instead of a complement — the two values can differ substantially, especially when is large.
C4 — Mean and Standard Deviation
Instead of computing over all terms, we can use shortcut formulas derived from the binomial PMF.
Binomial Mean and Standard Deviation
For :
These are special cases of the general L-PR-4 formulas — they produce the same numbers you would get by summing over all values, but without the algebra.
Mini-example — mean and SD for the free-throw player.
.
In the long run, she makes an average of 6 free throws per game. The number made typically deviates from 6 by about 1.22 free throws.
, NOT and NOT (that is the variance, not the SD). Forgetting to take the square root — or taking the square root of only part of the expression — is the most common computational error in this concept.
C5 — Shape of the Binomial Distribution
The shape of depends on the value of :
: The distribution is degenerate — all probability is at .
: The distribution is right-skewed — most probability mass is near low values of , with a long right tail. The skew becomes more pronounced as approaches 0 for a fixed .
: The distribution is perfectly symmetric — the bar chart is a mirror image around .
: The distribution is left-skewed — most mass is near high values of , with a long left tail.
: The distribution is degenerate — all probability is at .
As increases with a fixed : The magnitude of the skewness, , decreases. A bell-shaped Normal approximation becomes reasonable when both and are sufficiently large.
Use the interactive visualization below to explore these relationships directly. Start with , and observe the right skew. Then increase toward 0.70 and watch the distribution flip to left-skewed. Then double and observe how the shape becomes more symmetric.
n = 10, p = 0.30, μ = 3.00, σ ≈ 1.45, max P ≈ 0.27 — shape: right-skewed
Figure 3: Binomial PMF — P(X = k) for each value of k.
Adjust n and p with the sliders to explore how the shape, mean, and spread change.
Shape rule summary:
→ all probability at
→ right-skewed (tail points right)
→ symmetric
→ left-skewed (tail points left)
→ all probability at
For fixed , larger reduces the magnitude of the skew; a bell shape requires both and to be sufficiently large
C6 — Recognizing Non-Binomial Settings
Not every “count of successes” scenario is binomial. Three common cases where the binomial model fails:
Non-Binomial Settings
Sampling without replacement from a small population (violates I): If you draw 5 cards from a deck of 52 without replacing them, the probability of drawing a heart changes after each draw. Trials are dependent. Use the hypergeometric distribution instead.
Probability changes between trials (violates S): If a student’s probability of answering correctly increases as the quiz proceeds (learning effect), is not constant. The binomial formula does not apply; you would need the general PMF approach from L-PR-4.
No fixed number of trials (violates N): “Count the number of attempts until the first success” has no predetermined . Use the geometric distribution instead.
Scenario:
A 75% free-throw shooter keeps attempting shots until she makes 3. Count
how many shots she takes.
Draw 5 cards from a standard 52-card deck without replacement. Count how
many are hearts.
Flip a fair coin 10 times. Count the number of heads.
A machine's defect rate drifts upward during a shift. Inspect 20 items in
sequence and count the defects.
Mark each BINS condition as passes or fails for this scenario, then press
Check.
Binary
— each trial has exactly two outcomes
Independent
— one trial's outcome doesn't affect another's
fixed Number
— the number of trials n is set in advance
constant Success probability
— p is the same on every trial
Figure 4: Check the four BINS conditions before reaching for the binomial model. Mark each
condition pass or fail, then check your call — if even one condition fails, the binomial formula does
not apply, and the verdict names the setting you are actually in.
Mini-example — spotting the violation.
You draw 4 cards from a standard 52-card deck without replacement. Let = number of aces drawn.
B: Ace (success) or not (failure). ✓
I: After drawing one ace, only 3 aces remain among 51 cards. .FAIL.
The binomial model does not apply. Use the hypergeometric distribution (or multiply conditional probabilities).
Sampling without replacement from a finite population is exactly hypergeometric, not binomial, because the draws are dependent. When the sampling fraction is small—roughly under the conservative rule used here— may be a useful approximation to a population with successes among members. A common error is to call that approximation exact merely because there are two outcomes.
Section 4: Worked Examples
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Example 1 — BINS Check and Exact Probability (Fully Worked)
Problem: A quality control inspector samples 6 light bulbs from a production line where each bulb is defective independently with probability 0.15. Let = number of defective bulbs in the sample. Find .
Step 1 — Check BINS.
I notice this is a sampling scenario and I need to check four conditions before applying any formula.
B: Each bulb is either defective (success for counting purposes) or not. ✓
I: The problem states “independently” — each bulb’s status does not depend on the others. ✓
N: is fixed in advance. ✓
S: on every bulb. ✓
All four hold: . I may proceed with the formula.
Step 2 — Identify parameters.
,,,.
I choose because the question asks for “exactly 2 defective.”
Step 3 — Apply the formula.
Interpretation: There is approximately a 17.6% chance that exactly 2 of the 6 bulbs are defective.
Example 2 — Cumulative Probabilities (“At Most” and “At Least”)
Problem: For from Example 1:
Find and list the values of included.
Find and choose either a direct sum or complement.
Before revealing the solution, record both probability setups and explain why your chosen method for part (b) is efficient.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
(a) — sum three terms.
Using the unrounded term values:
There is about a 95.3% chance that 2 or fewer bulbs are defective.
(b) — complement is easier.
The event includes . Rather than calculate five terms, use its complement:
Compare your setup with the solution: the crucial difference is that “at most 2” includes 2, whereas the complement of “at least 2” stops at 1.
Example 3 — Computing and , then Interpreting in Context
Problem: A greenhouse plants 20 seeds under identical conditions. Each seed germinates independently with probability 0.60. Let be the number that germinate.
Compute ,, and .
Interpret and in context.
Before revealing the solution, decide whether should be above or below 2.5 and justify your prediction from the formula.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
Why this method? The question asks for the center and spread of a binomial count, so the shortcuts and replace a full PMF table.
Verify BINS: B ✓ (germinates/does not), I ✓ (stated independent), N ✓ ( fixed), S ✓ ( constant).
So .
Interpretation:
: Across many groups of 20 seeds, an average of 12 seeds germinate.
: The germination count typically differs from 12 by about 2.19 seeds.
The prediction is confirmed because .
Example 4 — Transfer: When the Binomial Model Fails
A trainee draws 5 cards from a standard 52-card deck without replacement and lets be the number of aces drawn. The trainee writes:
“Each draw is ace or not ace, so .”
”.”
Identify the first faulty claim, name every BINS condition affected, choose the appropriate model, and compute the correct probability of exactly 2 aces using combinations.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
First faulty claim: Step 1. Binary outcomes and a fixed hold, but the draws are not independent. After an ace is drawn, the next ace probability changes from to . Thus I fails, and constant S fails with it.
Why this method? Sampling without replacement from a finite population calls for a hypergeometric count: choose the wanted aces and wanted non-aces, then divide by all 5-card hands.
The trainee’s binomial calculation is about 0.0465. It is not merely rounded differently; it comes from a model whose constant-probability assumption is false.
Section 5: Guided Practice
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Problem 1 — BINS Classification (C1 + C6)
Each scenario is visible because the goal is a distinct discrimination, not cycling through a hidden bank.
Scenario A. A fair coin is flipped independently 10 times. Let be the number of heads. Which model is appropriate?
Scenario B. A bag contains 5 red and 3 blue marbles. Three marbles are drawn without replacement. Let be the number of red marbles drawn. Which model is appropriate?
Scenario C. Patients respond independently to a treatment with probability 0.30. A hospital admits patients until 3 respond. Let be the number of patients admitted. Which model is appropriate?
Problem 2 — Exact Binomial Probability (C1 + C2)
Problem 3 — Cumulative Binomial Probability (C1 + C3)
Problem 4 — Centre, Spread, and Shape (C4 + C5)
Section 6: Independent Practice
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Problem 1 — Full Binomial Workflow (C1 + C2 + C3 + C4)
Problem 2 — Cumulative Boundaries (C3)
Translate the boundary language before calculating; this solo-first problem can reveal the individual commits if you need them.
Problem 3 — Majority of Trials (C2 + C3)
A majority means more than half the trials are successes. First commit to which counts qualify, then compute the total probability.
Problem 4 — Diagnose a BINS Error (C1 + C6)
Read the student’s work, identify the first false claim, and then use the feedback to name the appropriate model.
Problem 5 — Centre, Spread, and Shape (C4 + C5)
Mixed Review — Retrieval from Earlier Lessons
These problems draw on concepts from earlier in the course. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.
Review Problem 1 — Combinations in Probability
Review Problem 2 — Expected Value and Variance
Section 7: Mastery Check
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Begin each item on your own. Feedback and model answers come only after your attempt.
Question 1 — Feynman Test
Explain in your own words why the binomial probability formula includes a combination coefficient . Use trials and successes as a concrete example, and write as if teaching a classmate who missed the lesson. Afterward, compare your explanation with the model.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Compare with a model answer
The combination coefficient is needed because there are multiple distinct arrangements of successes among trials, and each arrangement contributes the same probability.
Concrete example (,): Suppose each trial independently succeeds with probability . The sequences that produce exactly 2 successes are:
Sequence
Probability
S S F
S F S
F S S
Each sequence has the same probability . There are such sequences (choose which 2 of the 3 positions are successes). The total probability of exactly 2 successes is .
Without , the formula gives the probability of one specific sequence (e.g., SSF), not all sequences. The combination coefficient counts all arrangements and multiplies them together.
Question 2 — Applied Scenario
A seed treatment works for each seed independently with probability 0.85. A greenhouse treats 7 seeds.
(a) Let be the number of seeds for which the treatment works. Decide whether a binomial model is appropriate and state the distribution.
(b) Compute .
(c) Compute using an appropriate method. Show your setup.
Part (c1). Which direct-sum setup is correct?
Part (c2). Which complement setup is correct?
Full Solution
(a) BINS check: B ✓ (works/does not), I ✓ (stated independent), N ✓ ( fixed), S ✓ (). So .
(b)
(c) Two equivalent approaches:
Direct sum:
Using the unrounded term values:
Complement approach:.
Both give the same answer (0.7166). For this problem, the direct sum with 2 terms is simpler than summing 6 terms for .
Question 3 — Error Analysis
Find the first faulty step in the student’s binomial calculation.
Self-Assessment
How confident do you feel about the binomial distribution right now?
Still confusedGetting thereSolid graspFeeling strongReady for the Boss Fight
Section 8: Boss Fight
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Choose one path. Both require the same workload: compare two plans, decide which probability tools matter, quantify center and spread, and defend a recommendation.
⚙️ Path A — The Analyst
Compare two production lines and decide which one triggers an investigation under a defect-risk policy.
🎲 Path B — The Architect
Compare two outreach plans with the same expected response count and recommend the more reliable design.
⚙️ Path A — The Analyst
Scenario: A factory independently inspects 10 items from each of two continuous production lines. Line A has defect probability 0.10 per item; Line B has defect probability 0.20 per item. Policy requires an investigation when the chance of finding at least 2 defects in a 10-item inspection is 0.50 or greater.
Task 1. Define one random variable for each line, decide whether the same model family is justified for both, and support your decision with the scenario’s assumptions.
Task 2. For each line, determine the probability relevant to the policy. Choose an efficient setup, show it symbolically, and report each probability to four decimals.
Task 3. For each line, calculate the expected defect count and standard deviation, describe the distribution’s skew direction, and explain what the comparison reveals.
Task 4. Apply the policy and recommend which line or lines require investigation. Then state whether your model would remain defensible if the 10 items were drawn without replacement from a small fixed lot.
Reflection: Which decision mattered most to your conclusion—the model check, translating “at least 2,” choosing a complement, or interpreting the threshold? Explain why.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Complete Solution and Reflection Model
Task 1. Let and count defects in 10 inspections. Each item is defective or not (B), inspections are stated independent (I), 10 inspections are fixed (N), and each line has a constant defect probability (S).
Task 2. “At least 2” includes 2 through 10. The complement has only two terms:
For Line A:
For Line B:
Task 3. Line A:
Line B:
Both distributions are right-skewed because both success probabilities are below 0.5. Line B is centered at twice as many defects and has greater absolute spread.
Task 4. Line A does not meet the 0.50 investigation threshold because . Line B does because , so investigate Line B. If the items came without replacement from a small fixed lot, independence and constant success probability would fail; a hypergeometric model would be needed.
Reflection model: Translating “at least 2” was the decisive choice. Once I recognized that its complement is 0 or 1 defect, each calculation needed only two terms and the policy comparison became direct.
🎲 Path B — The Architect
Scenario: A community organization is choosing between two independently targeted outreach plans. A “response” means one contacted person signs up.
Compact plan: contact 5 people, each with response probability 0.40.
Broad plan: contact 10 people, each with response probability 0.20.
Both plans expect the same number of responses. The organization wants the plan with the greater chance of receiving at least 2 responses; if those chances are close, it prefers the plan with less variation.
Task 1. Define a random variable for each plan, decide whether the same model family is justified for both, and support your decision with the scenario’s assumptions.
Task 2. For each plan, determine the chance of meeting the response target. Choose an efficient setup, show it symbolically, and report each probability to four decimals.
Task 3. For each plan, calculate the expected response count and standard deviation, describe the distribution’s skew direction, and explain why equal means do not imply equal probability patterns.
Task 4. Recommend one plan using both the target probability and variability. Then identify one realistic change to the outreach process that would make the binomial model questionable.
Reflection: Which comparison was more informative here—equal means, different standard deviations, or the target-event probabilities? Explain what the other two still contributed.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Complete Solution and Reflection Model
Task 1. Let and count responses. Each contact responds or not (B), contacts are stated independent (I), the contact counts are fixed (N), and each plan has a constant response probability (S).
Task 2. For either plan:
For the Compact plan:
For the Broad plan:
Task 3. Compact plan:
Broad plan:
Both are right-skewed because . Equal means locate the same center, but the Broad plan spreads probability more widely across possible counts.
Task 4. Recommend the Compact plan: its target probability is higher () and its standard deviation is smaller (). The binomial model would become questionable if response probabilities changed substantially across contacts—for example, if later contacts received a redesigned message—or if one person’s response influenced another’s.
Reflection model: The target probability answers the decision directly, so it carries the recommendation. The equal means explain why the plans look comparable at first, while the standard deviations explain why their full probability patterns still differ.
Section 9: Challenge Problems
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Ready for more? These problems go beyond the lesson objectives.
Challenge 1 — Algebraic Derivation of
In L-PR-4, you learned . Show algebraically that this equals for a binomial random variable. Decide which algebraic rewrite makes the binomial sum collapse, and justify it.
Show Solution
Use the identity , which you can verify for small values (for example, ,). It rewrites the factor into the coefficient of a smaller binomial distribution.
The term is 0, so:
Apply the identity :
Factor out (since ):
Let , so as ranges from 1 to , ranges from 0 to :
The sum is the sum of all PMF values for , which equals 1:
Challenge 2 — The Mode of the Binomial Distribution
For , the mode is the count—or pair of counts—with the greatest probability. Predict the peak from and the shape before opening the generated comparison.
Section 10: Solutions Reference
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Full step-by-step solutions for all Worked Examples, Guided Practice, Independent Practice, Boss Fight, and Challenge Problems are available on the dedicated solutions page.