Lesson 12: Binomial Distributions

(a) Warm-Up

A basketball player makes 80% of her free throws. Tonight she shoots 10 of them. Before you do any math, answer these by gut feeling:

1. How many do you expect her to make?
2. Is it possible she makes all 10? Is it likely?
3. Which is more probable: making exactly 8, or making 8 or more?

Most people land on "about 8" for the first question — and that instinct is exactly right, because the expected count is μ = np = 10(0.80) = 8. The other two questions are harder to eyeball, and that's the point of this lesson. "Possible but not likely" needs a number attached to it.

In Lesson 11 you learned to build a probability distribution for any discrete random variable and compute its expected value and standard deviation by hand. That works, but it's slow. Today we meet a special, extremely common type of count — the binomial random variable — that shows up whenever you repeat the same success/failure trial a fixed number of times. It has a tidy formula and dedicated calculator commands, so you'll never have to build the whole table again.

(b) Core Concept

When is a count binomial? The BINS conditions

A binomial setting arises when you perform the same trial a fixed number of times and count how many times one particular outcome happens. To qualify, four conditions must hold. Remember them with BINS:

• B — Binary. Each trial has exactly two possible outcomes, which we label success and failure. ("Success" is just a label for the outcome we're counting — it doesn't mean "good.")
• I — Independent. The trials are independent: the result of one trial doesn't change the probabilities for any other.
• N — Number fixed. The number of trials, n, is set in advance.
• S — Same probability. Every trial has the same probability of success, p.

If all four hold, then X = the number of successes in n trials is a binomial random variable, written X ~ B(n, p). The possible values of X are 0, 1, 2, …, n.

Why "fixed n" matters so much: A binomial count fixes the number of trials and lets the number of successes vary. (If instead you fixed the number of successes and let the number of trials vary, you'd have a different setting entirely — one that is not part of this course. Always confirm n is decided up front.)

Checking BINS on a concrete example

A pharmacy fills prescriptions, and historically 30% of customers also buy an over-the-counter item at checkout. We watch the next 8 customers and count how many buy an OTC item. Is this binomial?

• B: Each customer either buys an OTC item (success) or doesn't (failure). ✓ Two outcomes.
• I: One customer's decision doesn't affect another's. We'll treat them as independent. ✓
• N: We fixed n = 8 in advance. ✓
• S: Each customer has the same probability p = 0.30. ✓

All four hold, so X ~ B(8, 0.30), where X = number of the 8 customers who buy an OTC item. We'll use this scenario throughout.

The 10% condition (sampling without replacement). Strictly, "independent" can fail when you sample without replacement from a finite population — each draw changes the next probability. In practice we treat trials as independent as long as the sample is less than 10% of the population (the 10% condition). Eight customers out of a large customer base easily satisfies this.

The binomial probability formula

The probability of getting exactly k successes in n trials is:

P(X = k) = C(n, k) · p^k · (1 − p)^(n − k)

Let's unpack each piece:

• C(n, k) — read "n choose k" — counts the number of different arrangements of k successes among n trials. It equals n! / (k! · (n − k)!). On the TI-84 it's n MATH → PRB → nCr → k.
• p^k — the probability of the k successes happening (each with probability p, multiplied because independent).
• (1 − p)^(n − k) — the probability of the remaining n − k failures.

One full calculation by hand

Using our pharmacy scenario X ~ B(8, 0.30), find P(X = 2) — the probability that exactly 2 of the 8 customers buy an OTC item.

Plug into the formula with n = 8, k = 2, p = 0.30:

P(X = 2) = C(8, 2) · (0.30)^2 · (0.70)^6

Step 1 — the combination:

C(8, 2) = 8! / (2! · 6!) = (8 · 7) / (2 · 1) = 56 / 2 = 28

Step 2 — the success factor:

(0.30)^2 = 0.09

Step 3 — the failure factor:

(0.70)^6 = 0.117649

Step 4 — multiply:

P(X = 2) = 28 · 0.09 · 0.117649
         = 28 · 0.01058841
         = 0.29647548  ≈  0.2965

So there's about a 29.65% chance that exactly 2 of the 8 customers buy an OTC item.

Confirm with the calculator:

TI-84: 2nd → DISTR → binompdf
binompdf(8, 0.30, 2)
Output: 0.2964754752  ≈  0.2965  ✓

Hand calculation and calculator agree. The "pdf" in binompdf stands for probability density function — use it for exactly k.

Mean and standard deviation of a binomial

You could find μ and σ the long way (Lesson 11: build the table, then μ = Σ x·P(x) and σ = √(Σ (x − μ)² · P(x))). But a binomial has shortcut formulas — memorize these:

μ = np σ = √(np(1 − p))

For our pharmacy count X ~ B(8, 0.30):

μ = np = 8 · 0.30 = 2.4
σ = √(np(1 − p)) = √(8 · 0.30 · 0.70) = √1.68 = 1.2961

Interpretation (this is the part that scores points): If we repeatedly watched groups of 8 customers, we'd expect about 2.4 of them, on average, to buy an OTC item, and the count would typically vary from that mean by about 1.30 customers.

A one-line note on the normal approximation

When n is large and p isn't too extreme, the binomial distribution starts to look bell-shaped, and you can approximate it with a normal curve (the usual check is np ≥ 10 and n(1 − p) ≥ 10). We'll develop that in Lesson 13 — for this lesson, keep it pure binomial and use the exact formula and calculator.

(c) Worked Examples

Example 1 — Exactly k (easy)

Problem. A multiple-choice quiz has 5 questions, each with 4 choices. A student guesses every answer at random. Find the probability of getting exactly 3 correct.

Strategy. Guessing with 4 choices means p = 1/4 = 0.25 of a correct answer, n = 5, count correct answers. BINS holds (each question binary, independent guesses, fixed n = 5, same p). Want exactly k = 3 → binompdf.

Solution.

P(X = 3) = C(5, 3) · (0.25)^3 · (0.75)^2
         = 10 · 0.015625 · 0.5625
         = 0.087890625  ≈  0.0879
TI-84: binompdf(5, 0.25, 3) = 0.0878906...

Interpretation. A random guesser has about an 8.8% chance of getting exactly 3 of the 5 right.

Example 2 — At most k, and at least k (medium)

Problem. Back to the free-throw shooter: she makes 80% of her shots and takes n = 10, so X ~ B(10, 0.80).

(a) Find P(X ≤ 7), the probability she makes at most 7.

(b) Find P(X ≥ 9), the probability she makes at least 9.

Strategy. "At most 7" is a cumulative-from-the-bottom probability → use binomcdf directly. "At least 9" needs the complement because binomcdf only adds up to a value.

Solution (a). "At most 7" = P(X ≤ 7):

TI-84: 2nd → DISTR → binomcdf
binomcdf(10, 0.80, 7) = 0.3222004736  ≈  0.3222

The "cdf" (cumulative distribution function) adds up P(X = 0) + P(X = 1) + … + P(X = 7).

Solution (b). "At least 9" = P(X ≥ 9) = 1 − P(X ≤ 8). Because we want 9 and up, subtract everything below 9, i.e. up through 8:

P(X ≥ 9) = 1 − binomcdf(10, 0.80, 8)
         = 1 − 0.6241903616
         = 0.3758096384  ≈  0.3758

Interpretation. She makes at most 7 about 32.2% of the time, and at least 9 about 37.6% of the time. Notice she's more likely to make 9-or-10 than to make 7-or-fewer — that's what an 80% shooter over 10 shots looks like.

Example 3 — Between two values + mean/SD (AP-style)

Problem. A factory's bolts are defective 4% of the time, independently. An inspector pulls a box of n = 50 bolts. Let X = number defective, X ~ B(50, 0.04).

(a) Find P(2 ≤ X ≤ 5), the probability that between 2 and 5 (inclusive) are defective.

(b) Find and interpret the mean and standard deviation of X.

Strategy. "Between 2 and 5 inclusive" = P(X ≤ 5) − P(X ≤ 1). Subtract the cdf just below the lower bound. Then apply μ = np and σ = √(np(1−p)).

Solution (a).

P(2 ≤ X ≤ 5) = binomcdf(50, 0.04, 5) − binomcdf(50, 0.04, 1)
             = 0.9856... − 0.4005...
             = 0.5851...  ≈  0.5851

Solution (b).

μ = np = 50 · 0.04 = 2.0 defective bolts
σ = √(np(1 − p)) = √(50 · 0.04 · 0.96) = √1.92 = 1.3856

Interpretation. There's about a 58.1% chance the box has between 2 and 5 defective bolts. On average we'd expect 2 defective bolts per box of 50, give or take about 1.39 bolts.

Example 4 — Is it even binomial? (conceptual)

Problem. For each scenario, decide whether X is binomial. If not, name the broken condition.

(a) Deal 5 cards from a standard 52-card deck; X = number of hearts.

(b) Roll a fair die until you get a 6; X = number of rolls.

(c) Survey 20 randomly chosen students from a school of 2,000; X = number who walk to school, given 35% walk.

Solution.

• (a) Not binomial. Dealing without replacement from 52 cards breaks Independent / Same p — after each heart removed, P(next heart) changes, and 5 out of 52 is nearly 10% of the deck, so we can't wave it off with the 10% condition.
• (b) Not binomial. The number of trials is not fixed — you keep rolling until success, so N fails. (This "keep going until the first success" setting is not part of this course; binomial requires a fixed n.)
• (c) Binomial (approximately). Binary (walks / doesn't), fixed n = 20, same p = 0.35, and 20 is well under 10% of 2,000 so independence is fine. X ~ B(20, 0.35). ✓

(d) Common Mistakes

1. Off-by-one on the "at least" complement. For P(X ≥ k), students write 1 − binomcdf(n, p, k). Wrong — that subtracts too much. To keep k itself, subtract everything below k: P(X ≥ k) = 1 − binomcdf(n, p, k − 1). Example: P(X ≥ 9) = 1 − binomcdf(n, p, 8), not …, 9).

2. Using binompdf when you need binomcdf (or vice versa). binompdf gives exactly k — a single bar. binomcdf gives k or fewer — a running total. "At most / no more than" → cdf. "Exactly" → pdf. Reaching for pdf on an "at most 7" problem (only counting X = 7) is a classic lost-points error.

3. Ignoring independence or a non-fixed n. Before any formula, run BINS. Sampling without replacement from a small population (cards, a small club) breaks independence. "Until the first success" breaks fixed-n and is not binomial at all. Don't compute a binomial probability for a non-binomial setting.

4. Standard-deviation formula slips. The SD is σ = √(np(1 − p)) — note the square root over the whole product. Two frequent errors: forgetting the square root (that gives the variance np(1−p), not the SD), and writing √(np). Also don't confuse it with μ = np.

5. "Between" boundary errors. For P(a ≤ X ≤ b) use binomcdf(n, p, b) − binomcdf(n, p, a − 1). Subtracting binomcdf(…, a) instead of binomcdf(…, a − 1) wrongly drops the value X = a.

(e) Practice Problems

Multiple choice — 4 choices each. Use a TI-84 where helpful.

Short answer.

A quality engineer inspects 25 microchips, each defective independently with probability 0.06. Let X = number defective.

(a) Verify the binomial setting (BINS).

(b) Find P(X = 0).

(c) Find P(X ≥ 3).

(d) Find μ and σ, and interpret μ in context.

A survey finds 45% of teens use a particular app. In a random sample of 16 teens, let X = number who use it.

(a) Find P(X = 8).

(b) Find P(X ≤ 6).

(c) Find P(6 ≤ X ≤ 10).

(d) Find the mean and standard deviation of X.

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## (f) FRQ Practice (10 points · Analyze Data)

The problem.

A regional airline knows from years of records that 18% of booked passengers fail to show up for a particular commuter flight ("no-shows"), and that passengers' show-up behaviors are independent of one another. For a flight with 20 booked passengers, let X = the number of no-shows.

(a) Verify that X can be modeled as a binomial random variable, and state the values of n and p. (2 points)

(b) Find the probability that exactly 4 passengers are no-shows. Show the calculator command you use. (2 points)

(c) Find the probability that at least 5 passengers are no-shows. (2 points)

(d) Find the mean and standard deviation of X, and interpret the mean in context. (3 points)

(e) The plane has 18 seats. The airline "overbooks" by selling all 20 tickets, gambling that enough people won't show. Using your work above, briefly comment on whether selling 20 tickets is a reasonable gamble. (1 point)

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### Model Response

(a) This is a binomial setting because the BINS conditions hold:

- Binary: each passenger either is a no-show (success) or shows up (failure).

- Independent: the problem states passengers' behaviors are independent.

- Number fixed: n = 20 booked passengers, fixed in advance.

- Same p: each passenger has probability p = 0.18 of being a no-show.

So X ~ B(20, 0.18) with n = 20 and p = 0.18.

(b) Want P(X = 4) — exactly 4, so binompdf:

P(X = 4) = binompdf(20, 0.18, 4) = 0.2153  (≈ 0.215)

(By hand: C(20,4)·(0.18)^4·(0.82)^16 = 4845 · 0.00104976 · 0.04230... ≈ 0.2153.)

(c) Want P(X ≥ 5). Use the complement, subtracting everything up through 4:

P(X ≥ 5) = 1 − binomcdf(20, 0.18, 4)
         = 1 − 0.7151
         = 0.2849  (≈ 0.285)

(d)

μ = np = 20 · 0.18 = 3.6 no-shows
σ = √(np(1 − p)) = √(20 · 0.18 · 0.82) = √2.952 = 1.718

Interpretation: Over many flights of 20 booked passengers, the airline expects an average of about 3.6 no-shows per flight, typically varying by about 1.72 no-shows.

(e) With 18 seats, all 20 booked passengers can fly only if there are at least 2 no-shows. P(X ≥ 2) = 1 − binomcdf(20, 0.18, 1) = 1 − 0.1018 = 0.898 — about a 90% chance enough seats free up, and the mean (3.6) comfortably exceeds the 2 no-shows needed. Selling 20 tickets is a reasonable gamble, though roughly 10% of the time too many passengers show and someone gets bumped.

Scoring Rubric (10 points)

Total: 10 points.

Where students lose points

• Part (a): Listing BINS as bare letters without connecting each to this context. The reader wants "binary — no-show or shows up," not just "B is satisfied."
• Part (c): The off-by-one error — writing 1 − binomcdf(20, 0.18, 5), which drops X = 5 and gives the wrong probability. For "at least 5," subtract through 4.
• Part (d): Forgetting the square root and reporting σ = 2.952 (that's the variance), or giving μ and σ but skipping the in-context interpretation — that interpretation is its own point.
• Part (e): Answering "yes/no" with no link to the probability or the mean. A judgment must be supported by a number.