Binomial Distribution

Section 1: Introduction

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Rapid COVID-19 tests have a sensitivity of about 70%: if you have COVID, the test correctly detects it 70% of the time. They also have a specificity of about 98%: if you don’t have COVID, the test correctly gives a negative result 98% of the time.

Suppose your workplace tests 200 employees, and 10 of them actually have COVID. How many employees will test positive? How many positive results will be false positives? How many infected employees will get a false negative and return to work unknowingly?

These aren’t abstract questions — they determine whether mass testing is effective or misleading. And answering them requires a model for counting successes and failures across many independent trials: the binomial distribution.

After this lesson, you will be able to:

By the end of this lesson you will be able to:

Check whether a situation satisfies the four BINS conditions for a binomial setting.
Apply the formula to compute exact binomial probabilities.
Compute cumulative probabilities for “at most ,” “at least ,” “more than ,” and “fewer than .”
Calculate and interpret the mean and standard deviation of a binomial random variable.
Identify situations where the binomial model does NOT apply and name the appropriate alternative.

The binomial distribution is one of the most useful models in statistics. Anywhere you have a fixed number of independent trials, each with the same probability of success — quality control, clinical trials, polling, genetics — the binomial applies.

Section 2: Prerequisites

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What you need coming into this lesson:

PR-4 — PMF validity: A probability mass function must satisfy and for all . Today’s formula is a special PMF, so these rules still apply.
PR-4 — Expected value: . The binomial shortcut is derived from exactly this formula — you do not need to re-derive it, but knowing where it comes from prevents it from feeling like magic.
PR-4 — CDF language: “At most ” means ; “at least ” means . This lesson extends that language with “more than” and “fewer than,” which shift the boundary by 1.
PR-3 — Combination formula: counts the number of ways to choose objects from without regard to order. This coefficient is the core of the binomial formula.
PR-3 — Distinguishing permutations from combinations: The binomial formula uses combinations (order does not matter — it doesn’t matter which 3 patients improve, only that exactly 3 do).

Retrieval Checkpoint — warm up the machinery you’ll need.

Problem 1. Compute .

I can verify that a two-row PMF sums to 1 (PR-4, Section 3). I know the difference between

and

(PR-4, Section 3). I can compute

for small values by hand (PR-3, Section 3).

Success Factor:

Bridge — where PR-4 ends and PR-5 begins.

In PR-4, every probability distribution was custom-built: you listed outcomes and assigned probabilities from first principles. From this lesson onward, you are working with a named family of distributions. The binomial PMF gives you a formula so that you never have to list all sequences of successes and failures. The formula is valid only when four specific conditions hold — checking those conditions (BINS) is always Step 1.

Retrieval Warm-up — from earlier lessons

A discrete random variable has the PMF below. What is ?

	0	1	2	3
	0.20	0.35	0.30	0.15

A discrete random variable has PMF:

	0	1	2	3
	0.10	0.40	0.35	0.15

What is ?

Section 3: Core Concepts

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Navigation — six concepts in this section:

C1 — BINS Conditions: The four requirements that define a binomial setting.
C2 — Binomial Probability Formula: .
C3 — Cumulative Binomial Probabilities: “At most,” “at least,” “more than,” “fewer than.”
C4 — Mean and Standard Deviation: ; .
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.

A rapid test for strep throat correctly detects an infection 85% of the time. Suppose a clinic tests 20 patients who actually have strep. How many positive results do they expect?

The naive answer is . But “17 positive results” is just the average over many repetitions of this experiment. In any one run of 20 tests, you might get 14 positive results, or 18, or 16. The question is: what’s the complete picture of all possible outcomes?

Think of it this way: each test is an independent coin flip (heads = positive, tails = false negative) with . Running 20 such flips gives you a count of successes. That count varies from trial to trial. The pattern of that variation — which counts are likely and which are unlikely — is the binomial distribution.

In 10,000 simulated runs of this experiment:

About 7% get exactly 17 positive results (the most likely single outcome)
About 97% of runs produce between 13 and 20 positive results
Getting fewer than 10 positive results would be extraordinary (probability < 0.001)

The binomial distribution gives us the exact probability for every possible count. This is more useful than just knowing the average — it tells clinics 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 geometric setting.

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.

Success probability p

p = 0.40, p²(1−p) = 0.0960, P(X = 2) = 0.2880

Figure: 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.

Boundary value k = 3

P(X ≤ 3) = 0.5941 for X ∼ B(8, 0.40)

Figure: X ∼ B(8, 0.40) PMF. Orange bars are included in the selected probability; grey bars are excluded. The blue 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 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 perfectly symmetric — the bar chart is a mirror image around .
: The distribution is right-skewed — most probability mass is near low values of , with a long right tail. The most extreme case is close to 0.
: The distribution is left-skewed — most mass is near high values of , with a long left tail.
As increases (holding fixed): The distribution becomes more bell-shaped (less skewed) regardless of . This foreshadows the Normal approximation in PR-7.

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.

Number of trials n Success probability p

n = 10, p = 0.30, μ = 3.00, σ = 1.45

Figure: 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:

→ right-skewed (tail points right)
→ symmetric
→ left-skewed (tail points left)
Large → more bell-shaped for any

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 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.

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” is not binomial unless the population is very large relative to the sample (roughly is the rule of thumb). A common error is to apply the binomial formula to card-drawing or urn-without-replacement problems just because there are two outcomes.

Section 4: Worked Examples

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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.

P(X = 2) = inom{6}{2}(0.15)^2(0.85)^4

= 15 imes 0.011745 pprox \mathbf{0.1762}

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 (a) and (b) .

Prediction checkpoint — before you see the solution:

Part (a) asks for “at most 2.” Which values of are included? (0, 1, 2 — three terms to sum.)

Part (b) asks for “at least 2.” Would you sum from 2 to 6 (five terms), or use the complement? Which approach is easier?

Show Solution

(a) — sum three terms.

P(X = 0) = inom{6}{0}(0.15)^0(0.85)^6 = 1 imes 1 imes 0.3771 = 0.3771

P(X = 1) = inom{6}{1}(0.15)^1(0.85)^5 = 6 imes 0.15 imes 0.4437 = 0.3993

There is about an 85.3% chance that 2 or fewer bulbs are defective.

(b) — complement is easier.

“At least 2” = .

Example 3 — Computing and , then Interpreting in Context

Problem: A new medication is effective for each patient independently with probability 0.60. In a clinical trial with patients, let = number for whom the medication is effective. Compute and , and interpret both in context.

Show Solution

Verify BINS: B ✓ (effective/not), I ✓ (stated independent), N ✓ ( fixed), S ✓ ( constant).

So .

Interpretation:

: In repeated clinical trials of 20 patients each, the medication is effective for an average of 12 patients per trial.
: The number of effective outcomes typically deviates from 12 by about 2.19 patients. A result of 9 or 15 effective patients would be unusual but not extreme (roughly 1.4 standard deviations from the mean).

Example 4 — Returning to the Introduction: COVID Testing at Scale

Recall the scenario from Section 1. A rapid COVID-19 test has 70% sensitivity — it correctly detects an infection 70% of the time. Your workplace tests 10 employees who actually have COVID. Let = the number who receive a correct positive result.

Show Solution

Step 1 — Verify BINS:

B — Binary outcomes: Each result is either a correct positive (success) or a false negative (failure). ✓
I — Independent trials: Each employee’s test result is independent of the others. ✓
N — Fixed number of trials: infected employees are tested. ✓
S — Constant success probability: on every test. ✓

So .

Step 2 — Exactly 7 correctly detected:

There is approximately a 26.7% chance that exactly 7 of the 10 infected employees test positive.

Step 3 — At least 8 correctly detected: (three terms is manageable; sum directly):

Step 4 — Mean and standard deviation:

Interpretation: Among 10 infected employees, the test correctly identifies an average of 7. The number detected typically deviates from 7 by about 1.45 employees. Even in a best-case run, roughly 3 infected employees are expected to slip through as false negatives — returning to work unknowingly. This is the concrete cost of a 70% sensitivity in practice.

Section 5: Guided Practice

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Problem 1 — BINS Classification (Variant Bank)

For each problem set below, classify each scenario as binomial or not binomial. If not binomial, state which BINS condition fails and name the appropriate alternative model if one applies.

Set A. Classify each:

A coin is flipped 10 times. = number of heads.
A student randomly selects answers on a 10-question true/false test. = number correct.
A bag contains 5 red and 3 blue marbles. Three marbles are drawn without replacement. = number of red marbles drawn.