IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 ~7 min read

Conditional Probability

Conditional probability is the chance of A happening given that B has already happened — written P(A | B). It shows up everywhere in IB exams: drawing without replacement, two-stage problems, frequency tables, and the foundation for Bayes’ theorem. The formula itself is simple — what’s tricky is choosing the right denominator.

📘 What you need to know

Notation: P(A | B) reads “probability of A given B“.
Formula: P(A | B) = P(A ∩ B)P(B) (needs P(B) ≠ 0).
Rearranged: P(A ∩ B) = P(B) · P(A | B) = P(A) · P(B | A).
Without replacement ⟹ second draw uses conditional probability (the bag has changed).
Independence test: P(A | B) = P(A) means A and B are independent.
Frequency tables: P(A | B) = (count in both A and B) ÷ (count in B).
Not symmetric: P(A | B) ≠ P(B | A) in general.
Reduced sample space: the denominator P(B) is your “new universe” — only count outcomes where B is true.

The conditional probability formula

Conditional probability P(A | B) = P(A ∩ B)P(B)

Read the right-hand side as: out of the times B happens, what fraction also have A. That fraction is your conditional probability. The denominator P(B) is your new sample space — you’ve zoomed in on a smaller world.

Rearranged — find the intersection P(A ∩ B) = P(B) · P(A | B) = P(A) · P(B | A)

The two rearranged versions are interchangeable — pick whichever pair (P(A), P(B | A)) or (P(B), P(A | B)) the question gives you.

Two situations where it shows up

Without replacement

P(2nd | 1st)

probabilities change after the first draw — count what’s left

Given information

P(A | B)

a fact narrows the sample space — recompute over B only

Spotting cue: if a question says “given that…”, “of those who…”, or describes drawing/picking without replacement, you’re in conditional-probability territory. Reach for the formula or count directly within the reduced sample space.

Linking conditional probability to independence

Independence — conditional version A and B independent ⟺ P(A | B) = P(A)

If knowing B happened does not change the probability of A, the events are independent. Either side of the iff works as a test — compute P(A | B) and compare to P(A); if equal, they’re independent, otherwise not.

🧭 Recipe — finding a conditional probability

Identify A and B: which event is the “given” (the condition), which is the one you’re finding.
Find P(B) — this becomes your denominator (the reduced sample space).
Find P(A ∩ B) — given directly, or by counting outcomes in both A and B.
Divide: P(A | B) = P(A ∩ B) ÷ P(B).
Sense-check: result must lie between 0 and 1.

Worked examples

WE 1

Conditional probability by counting (without replacement)

A bag contains 4 green marbles and 6 yellow marbles. Two marbles are drawn one after the other without replacement. Given that the first marble drawn is green, find the probability that the second is also green.

After 1st green is removed: 3 green and 6 yellow left Total marbles remaining = 9 Apply conditional probability by direct counting P(2nd green | 1st green) = 3/9 = 1/3 P(2nd green | 1st green) = 1/3 no formula needed — just recount what’s in the bag after the first event

WE 2

Direct application of the conditional formula

A factory has two production lines. The probability a randomly chosen product is defective AND from line X is 0.12. The probability a product comes from line X is 0.30. Find the probability that a product is defective given it came from line X.

Identify the formula P(D | X) = P(D ∩ X) / P(X) Substitute P(D | X) = 0.12 / 0.30 = 0.4 P(defective | line X) = 0.4 “given it came from line X” — line X is your reduced sample space

WE 3

Find P(A ∩ B) using the rearranged formula

At a music school, the probability a student plays piano is 0.40. Given that a student plays piano, the probability they also play guitar is 0.25. Find the probability that a randomly chosen student plays both piano and guitar.

Use the rearranged formula P(P ∩ G) = P(P) · P(G | P) Substitute = 0.40 × 0.25 = 0.10 P(plays both) = 0.10 when given P(A) and P(B|A), multiply directly — this is the engine of every tree diagram

WE 4

Find P(B) by rearranging the formula

In a clothing store, the probability that a customer buys something given that they pick up a basket is 0.75. The probability that a customer both picks up a basket and buys something is 0.45. Find the probability that a customer picks up a basket.

Let A = picks up basket, B = buys something P(B | A) = P(A ∩ B) / P(A) Substitute and rearrange for P(A) 0.75 = 0.45 / P(A) P(A) = 0.45 / 0.75 = 0.6 P(picks up basket) = 0.6 label A and B early — it stops you mixing up which probability is the denominator

WE 5

Test independence using conditional probability

For events G and H: P(G) = 0.30, P(H) = 0.50, P(G ∩ H) = 0.18. Use conditional probability to determine whether G and H are independent.

Compute P(G | H) P(G | H) = P(G ∩ H) / P(H) = 0.18 / 0.50 = 0.36 Compare with P(G) P(G | H) = 0.36, P(G) = 0.30 0.36 ≠ 0.30 → knowing H changes the probability of G G and H are NOT independent always state both numbers and the conclusion — examiners want the comparison shown

WE 6

Reversing the conditional from a frequency table

A health club surveys 200 members. The results are split as follows:

	Owns fitness tracker	No tracker	Total
Runner	50	30	80
Non-runner	30	90	120
Total	80	120	200

Given a member owns a fitness tracker, find the probability they are a runner.

Reduce sample space to tracker-owners only Total tracker-owners = 50 + 30 = 80 Count runners within the tracker-owners Runners with tracker = 50 Apply conditional formula directly from counts P(R | F) = 50 / 80 = 5/8 = 0.625 P(runner | owns tracker) = 5/8 = 0.625 on a frequency table: numerator = both, denominator = the “given” column total

💡 Top tips

Translate “given” to “/”: “given B” sits to the right of the bar, and B goes in the denominator.
“Of those who…” is a conditional — same skill, different wording. Restrict to that group as your denominator.
Frequency tables shortcut: count outcomes in both events for the numerator, count outcomes in the “given” event for the denominator. No formula needed.
Without replacement: don’t pre-compute — recount what’s left in the bag after each draw.
P(A | B) and P(B | A) are different — never swap them. Bayes’ theorem (next note) tells you how to convert.

⚠ Common mistakes

Using P(A ∩ B) when the question wants P(A | B) — divide by P(B) to get the conditional.
Wrong denominator — the “given” event always goes underneath, never the event you’re finding.
Treating P(A | B) as P(B | A) — they are usually different and may differ wildly.
Forgetting that without-replacement events are not independent — you must condition on the first draw.
Computing on the full sample space when the question reduces it — “of those who own a tracker” means the tracker-owners are your new total.

Next: Bayes’ Theorem (HL only). The formal way to reverse a conditional probability — given P(A | B), how do you find P(B | A)? It’s a Paper 1 favourite for HL students.

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