IB Maths AI SLBinomial DistributionPaper 1 & 2X ~ B(n, p)~7 min read
The Binomial Distribution
A binomial distribution counts the successes in a fixed number of repeated trials — free throws made out of 20, defective items in a batch of 50. If four conditions hold, the situation is captured by two numbers, X ~ B(n, p), with ready-made formulas for the mean and variance.
📘 What you need to know
Binomial = counting successes in n fixed trials, written X ~ B(n, p) — n trials, probability p of success each time.
The four conditions: a fixed number of trials, independent trials, exactly two outcomes (success / failure), and a constant probability p. All four must hold.
Probability of r successes: P(X = r) = nCrpr(1−p)n−r. In the booklet — the GDC does it for you.
Mean: E(X) = np — the expected number of successes. In the booklet.
Variance: Var(X) = np(1−p); the standard deviation is its square root. In the booklet.
Shape: p < 0.5 ⇒ tail to the right · p = 0.5 ⇒ symmetric · p > 0.5 ⇒ tail to the left.
Recognising a binomial distribution
A binomial distribution arises when an experiment is repeated and you count the successes — but only if all four conditions below hold.
The four conditions — FITC: Fixed number of trials n · Independent trials · Two outcomes only (success / failure) · Constant probability p. Miss one and it is not binomial.
If all four hold, write X ~ B(n, p): ~ means “is distributed as”, B is binomial, n the number of trials, p the probability of success. The probability of failure is 1 − p, sometimes written q.
Binomial probability formula
P(X = r) = nCr × pr × (1 − p)n−rprobability of exactly r successes in n trials, for r = 0, 1, 2, …, n
“Success” is just a label — a defective bulb or a late parcel can be the “success” you count.
Mean, variance and shape
Because a binomial is fully described by n and p, its average and spread come straight from formulas — no probability table needed.
Mean and variance of a binomial distribution
E(X) = npVar(X) = np(1 − p)standard deviation = √Var(X) — both formulas are in the booklet
The shape depends entirely on p. Drawn as a vertical line graph, it leans one way or the other:
All three have n = 10 — only p changes. A small p leans the graph low with a right tail; p = 0.5 is symmetric; a large p leans it high with a left tail.
Setting up a binomial model
To model a real situation, name the parts: identify what one trial is, decide what counts as a success, then read off n and p. Always define your variable in words before writing X ~ B(n, p).
Equally important: knowing when the model fails. A situation is not binomial if:
Not binomial when — the number of trials is not fixed (emails received in an hour) · trials affect each other (drawing cards without replacement) · there are more than two outcomes (a person’s shoe size) · the probability changes (a swimmer tiring over 50 lengths).
One exception: a random sample from a large population is technically “without replacement”, but the probability barely changes — so a binomial model is a good approximation.
🧭 Recipe — any binomial distribution question
Identify the trial and the success: what is one repetition, and which outcome are you counting? Success is just a label.
Check the four conditions (FITC): fixed n, independent, two outcomes, constant p. If any fails, it is not binomial.
State the model: write X ~ B(n, p) and say in words what X counts.
Pick the right tool: one exact value → the P(X = r) formula · mean → E(X) = np · spread → Var(X) = np(1−p).
Answer in context: give the result with its meaning (“about 18 calls”), not just a bare number.
Worked examples
WE 1
Check the four conditions
A 10-question multiple-choice quiz gives 5 options per question. A student answers every question by guessing. Let X be the number of correct answers. Show that X is binomial and state its distribution.
test the four conditions (FITC)Fixed: n = 10 questions ✓Independent: one guess doesn’t affect another ✓Two outcomes: correct or wrong ✓Constant p: each guess p = 1/5 = 0.2 ✓all four hold ⇒ binomialX ~ B(10, 0.2)With 5 equally likely options, p = 1/5 = 0.2. All four conditions pass, so the binomial model fits.
WE 2
Spot why a model is NOT binomial
A bag contains 7 red and 5 blue marbles. Four marbles are drawn one at a time without replacement. Let X be the number of red marbles drawn. Explain why X is not binomial.
check the probability of success each drawdraw 1: P(red) = 7/12if a red is taken, draw 2: P(red) = 6/11the probability of success CHANGEStrials are also not independent (no replacement)NOT binomial — p is not constantWithout replacement, each draw changes the bag — two conditions fail at once: p is not constant, and trials are not independent.
WE 3
Set up a model and find E(X) = np
At a call centre, 12% of all calls are about billing. During one shift, 150 calls are received. Let B be the number of billing calls. (a) State the distribution of B. (b) Find the expected number of billing calls.
(a) identify the trial, n and ptrial = one call · success = a billing calln = 150, p = 0.12B ~ B(150, 0.12)(b) expected value: E(X) = npE(B) = 150 × 0.12 = 18(a) B ~ B(150, 0.12) · (b) E(B) = 18 callsE(X) = np is in the booklet — the mean of a binomial is just trials × probability, no table needed.
WE 4
Variance and standard deviation
A courier finds that 35% of parcels arrive within 24 hours. In a random sample of 80 parcels, let X be the number that arrive within 24 hours. Find (a) E(X), (b) Var(X), (c) the standard deviation of X.
model: X ~ B(80, 0.35), so n = 80, p = 0.35(a) E(X) = np= 80 × 0.35 = 28(b) Var(X) = np(1 − p)= 80 × 0.35 × 0.65 = 18.2(c) standard deviation = √Var(X)= √18.2 ≈ 4.27(a) E(X) = 28 · (b) Var(X) = 18.2 · (c) SD ≈ 4.27Var(X) uses 1 − p = 0.65, the failure probability. The SD is always its square root — don’t stop at the variance.
WE 5
Use the formula for P(X = r)
A spinner has 6 equal sectors, exactly one coloured gold. It is spun 9 times. Let G be the number of gold results. Use the binomial formula to find P(G = 2).
model: G ~ B(9, 1/6)n = 9, p = 1/6, r = 2apply P(X=r) = nCr · pr · (1−p)n−rP(G=2) = 9C2 × (1/6)2 × (5/6)7≈ 36 × 0.02778 × 0.27908P(G = 2) ≈ 0.279The formula needs p2 for the 2 successes and (5/6)7 for the 7 failures — the powers must add to n = 9.
WE 6
Full model: sample, mean, spread and shape
In a large town, 60% of households own a pet. A researcher randomly samples 25 households. Let X be the number that own a pet. (a) State the distribution and the assumption needed. (b) Find E(X). (c) Find the standard deviation. (d) State, with a reason, whether the line graph of X has a tail to the left or right.
(a) model the sampleX ~ B(25, 0.6)assume the town is large, so each household independently has p = 0.6(b) E(X) = np= 25 × 0.6 = 15(c) Var(X) = np(1−p) = 25 × 0.6 × 0.4 = 6SD = √6 ≈ 2.45(d) the shape depends on pp = 0.6 > 0.5 ⇒ tail to the LEFT(a) X ~ B(25, 0.6) · (b) 15 · (c) ≈ 2.45 · (d) tail leftA large population means removing 25 households barely shifts p, so binomial is fine. Since p > 0.5, outcomes pile up high — the tail is on the left.
💡 Top tips
Memorise FITC: Fixed trials, Independent, Two outcomes, Constant p. All four must hold before you write X ~ B(n, p).
“Success” is only a label — defective, late or immune can all be the success you count. Don’t be put off by a “bad” outcome.
Define X in words before stating the distribution — examiners award a mark for it.
E(X) = np and Var(X) = np(1−p) are in the booklet — know which one a question wants.
A large population means sampling without replacement can still be modelled binomially — p barely changes.
⚠ Common mistakes
Ignoring independence: drawing cards or marbles without replacement breaks the binomial conditions.
Changing probability: if p shifts between trials (tiredness, a learning effect), the model is not binomial.
More than two outcomes: a trial must reduce to success / failure — a dice roll has six outcomes, so it isn’t directly binomial.
Confusing Var(X) with the standard deviation: Var(X) = np(1−p); the SD is its square root.
Swapping p and 1−p: the formula uses pr for successes and (1−p)n−r for failures — don’t reverse them.
Next up: Calculating Binomial Probabilities — using your GDC’s Binomial PD and CD functions for single probabilities P(X = x) and cumulative ones like P(X ≤ x) and P(a ≤ X ≤ b). The key skill is turning words — “at least”, “fewer than”, “more than” — into the right range of integers.
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