IB Maths AA SL Topic 4 — Stats & Probability Paper 2 🎯 Skill ~3 min practice

AA SL Binomial Probability skills

Binomial models the number of successes in n independent trials with probability p. The IB never asks you to compute these by hand — your GDC has dedicated functions. Master binomPdf and binomCdf and most binomial questions are 30-second wins.

The Method

X ~ B(n, p) n = number of trials · p = probability of success

Function 1

P(X = k)

binomPdf(n, p, k)

“exactly k” successes

Function 2

P(X ≤ k)

binomCdf(n, p, 0, k)

“at most” or “no more than”

Function 3

P(X ≥ k)

1 − binomCdf(…0, k−1)

“at least” — use complement

When does binomial actually apply?

Fixed number of trials (n). e.g. “12 free throws”, “20 customers”.
Each trial has only two outcomes — success or failure (no partial wins).
Constant probability of success (p) on every trial.
Trials are independent — one outcome doesn’t affect the next.

How to find binomPdf / binomCdf

TI-84 Plus

Press 2nd → VARS (DISTR menu)
Scroll to A: binompdf( or B: binomcdf(
Enter trials (n), p, and x value (or lower/upper)
Press ENTER — read off probability

Casio fx-CG50

Open STAT menu → DIST (F5)
Choose BINM (F5)
Pick Bpd (F1) for P(X=k) or Bcd (F2) for P(X≤k)
Set Data: Variable, then enter x, n, p

Worked examples

WE 1 EASY

A coin is biased so that P(heads) = 0.6. It is tossed 10 times. Find the probability of exactly 7 heads, to 3 sf.

step 1 — identify the binomial X ~ B(10, 0.6) — count heads in 10 tossesstep 2 — “exactly 7” → use binomPdf P(X = 7) = binomPdf(10, 0.6, 7)step 3 — read off GDC ≈ 0.21499…P(X = 7) ≈ 0.215 (3 sf) “exactly” is the keyword for binomPdf — single value, no range.

WE 2 MEDIUM

A basketball player has a free-throw success rate of 0.75. They take 12 shots. Find the probability that they make at most 8 shots, to 3 sf.

step 1 — set up the binomial X ~ B(12, 0.75)step 2 — “at most 8” means X ≤ 8 P(X ≤ 8) = binomCdf(12, 0.75, 0, 8)step 3 — GDC value ≈ 0.39068…P(X ≤ 8) ≈ 0.391 (3 sf) “at most k” = P(X ≤ k) → binomCdf from 0 to k.

WE 3 HARD

A factory produces light bulbs with a 4% defect rate. In a batch of 50, find the probability that at least 3 are defective, to 3 sf.

step 1 — binomial set-up X = number of defective X ~ B(50, 0.04)step 2 — “at least 3” → use complement P(X ≥ 3) = 1 − P(X ≤ 2)step 3 — GDC P(X ≤ 2) = binomCdf(50, 0.04, 0, 2) ≈ 0.67673 P(X ≥ 3) = 1 − 0.67673 = 0.32327P(X ≥ 3) ≈ 0.323 (3 sf) “at least k” → 1 − P(X ≤ k−1). don’t subtract from k itself!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Use your GDC for all of these — that’s the skill.

Q1 EASY A die is rolled 8 times. Find P(exactly 2 sixes), to 3 sf. Show answer ▼Hide answer ▲

X ~ B(8, 1/6) P(X = 2) = binomPdf(8, 1/6, 2) ≈ 0.260 (3 sf)

Q2 EASY A multiple-choice test has 15 questions, 4 options each (random guess). Find P(at most 3 correct), to 3 sf. Show answer ▼Hide answer ▲

X ~ B(15, 0.25) P(X ≤ 3) = binomCdf(15, 0.25, 0, 3) ≈ 0.461 (3 sf)

Q3 MEDIUM In a survey, 60% of people own a smartphone. From a sample of 20, find P(at least 15 own one), to 3 sf. Show answer ▼Hide answer ▲

X ~ B(20, 0.6) — use complement P(X ≥ 15) = 1 − P(X ≤ 14) = 1 − binomCdf(20, 0.6, 0, 14) = 1 − 0.8744 = 0.1256 ≈ 0.126 (3 sf)

Q4 MEDIUM A seed has 80% germination rate. 25 seeds are planted. Find P(between 18 and 22 inclusive germinate), to 3 sf. Show answer ▼Hide answer ▲

X ~ B(25, 0.8) — between two values P(18 ≤ X ≤ 22) = binomCdf(25, 0.8, 18, 22) ≈ 0.760 (3 sf) when “between a and b inclusive” — set lower = a, upper = b directly!

Q5 HARD A footballer scores from 35% of free kicks. In 14 attempts, find P(more than 5 goals), to 3 sf. Show answer ▼Hide answer ▲

“more than 5” means X ≥ 6 (not X ≥ 5) P(X > 5) = P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − binomCdf(14, 0.35, 0, 5) = 1 − 0.7218 = 0.2782 ≈ 0.278 (3 sf) “more than 5” excludes 5 itself! always check inclusive vs exclusive.

⚠ Common mistakes

Mixing up binomPdf and binomCdf. “Exactly” → Pdf. “At most” / “at least” → Cdf.
“At least k” → using P(X ≤ k) instead of P(X ≤ k−1). 1 − P(X ≤ k−1) is correct. Off-by-one is a one-mark loss.
Using the wrong probability of success. Re-read the question — “fails” or “doesn’t” might flip you to 1 − p.
Forgetting independence. If the situation says “without replacement”, binomial doesn’t apply — trials aren’t independent.
“More than k” vs “at least k”. “More than 5” is X ≥ 6. “At least 5” is X ≥ 5. The boundary case matters.
Rounding too early. Carry the GDC’s full value through any further steps — only round the final answer.

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Want the theory?

Read the full Binomial Distribution notes for the formula derivation, mean and variance (E(X) = np, Var(X) = np(1−p)), and how to spot a binomial situation in word problems.

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