IB Maths AA SL
Topic 4 — Probability
Paper 1 & 2
~10 min read
Conditional Probability
Sometimes new information changes a probability. If you know it’s raining, the chance of seeing an umbrella jumps. Conditional probability is the maths of “given that…” — finding the probability of one event when another has already happened.
📘 What you need to know
- Conditional probability P(A|B) means “the probability of A happening, given that B has already happened”.
- Formula: P(A|B) = P(A ∩ B)P(B) (in the formula booklet).
- Rearranged: P(A ∩ B) = P(B) × P(A|B) = P(A) × P(B|A).
- Independence test: if P(A|B) = P(A), the events are independent.
- “Without replacement” problems are classic conditional probability — the second draw depends on the first.
What does P(A|B) mean?
The notation P(A|B) reads as “the probability of A, given B“. The vertical bar is the key: it means “given that”.
The whole idea: knowing that B has happened often changes the probability of A. We’ve narrowed our world down to outcomes where B is true — and we want to find what fraction of those also have A.
🤔 An everyday example
The probability that a random person owns an umbrella is, say, 0.4. But if you tell me “given that it’s raining today”, that probability suddenly jumps to maybe 0.9 — people who carry umbrellas are way more visible in the rain.
Same person, same world — but new information narrows down the picture. That’s conditional probability.
The classic example — without replacement
Conditional probability shows up most often in “draw without replacement” questions:
Imagine a bag with 10 balls — 6 red and 4 blue. You draw two without replacing the first. What’s the probability that the second is red, given that the first was red?
Just count what’s left after the first draw
- Originally: 10 balls, 6 red, 4 blue.
- After drawing 1 red: 9 balls left, 5 of which are red.
- So P(2nd red | 1st red) = 59.
Notice how the original probability of red was 610 = 0.6, but conditional on a red being drawn first, it dropped to 59 ≈ 0.56. The first draw changed the second.
📍“Without replacement” → conditional probability
If you ever see “without replacement” in a question, the events are NOT independent — and you should expect to use conditional probability somewhere.
The conditional probability formula
Sometimes you can’t just count what’s left — you need a formula. Here it is:
In words: “the probability of A AND B both happening, divided by the probability of B happening”. The bottom — P(B) — represents the new “shrunken” sample space (only outcomes where B happened).
🧠Memory trick: “The given event goes on the bottom”
P(A|B) — B is the “given” (after the bar). It goes in the denominator. The intersection P(A ∩ B) goes on the top. So: P(A given B) = (overlap) ÷ (the given event).
Three useful forms of the same idea
All three formulas come from the same definition. Pick the one that lets you plug in the values you already have. If you have P(B) and P(A|B), use the first multiplication. If you have P(A) and P(B|A), use the second.
Using conditional probability to test independence
Here’s a clean way to spot independence using P(A|B):
The reasoning: independence means knowing B happened tells you nothing about A — so P(A|B) and P(A) must be equal.
By the same logic, all of these also work:
- P(A|B‘) = P(A)
- P(B|A) = P(B)
- P(B|A‘) = P(B)
And of course, the multiplication test still works too: P(A ∩ B) = P(A) × P(B) (from the previous note).
📍Two ways to test independence — pick the easier one
If you’ve got P(A ∩ B): use P(A ∩ B) = P(A) × P(B).
If you’ve got P(A|B): use P(A|B) = P(A).
Both work — pick whichever uses the values you already have.
The “shrunken sample space” trick
You don’t always need the formula. For “counting” problems, just shrink your sample space to only outcomes where B happened, then count how many of those also include A.
The shortcut method
- Restrict your sample space to outcomes where B is true. This is your new total.
- Count how many of those also satisfy A. This is your new “favourable” count.
- Divide the second by the first.
Example: out of 30 students, 18 study Spanish, and 8 of those 18 also study French. What’s P(French | Spanish)?
- Restrict to Spanish students: 18 of them.
- Of those, 8 also study French.
- So P(French | Spanish) = 818 = 49.
This shortcut is what the formula does behind the scenes — but with frequencies it’s faster and clearer. Use it whenever you can.
Worked examples
WE 1Find P(R) and test independence (Weatherville)
Let R = “it is raining in Weatherville” and T = “there is a thunderstorm in Weatherville”. It is known that P(T) = 0.035, P(T ∩ R) = 0.03 and P(T|R) = 0.15.
(a) Find P(R). (b) State whether R and T are independent. Justify your answer.
P(T) = 0.035, P(T ∩ R) = 0.03, P(T|R) = 0.15part (a) — find p(r)
Use formula: P(T|R) = P(T ∩ R)P(R)
Substitute values: 0.15 = 0.03P(R)
Rearrange: P(R) = 0.030.15 = 0.2
P(R) = 0.2part (b) — independence test
If R, T independent: P(T|R) = P(T).
Compare: P(T|R) = 0.15, P(T) = 0.035
0.15 ≠ 0.035
R and T are NOT independent (P(T|R) ≠ P(T))
always state the criterion AND show the numbers don’t match
WE 2Conditional probability without replacement
A bag contains 7 red and 3 blue balls. Two balls are drawn without replacement.
(a) Find the probability the second ball is blue, given that the first ball was red. (b) Find the probability that both balls are blue.
“Without replacement” → use the shrunken sample space.part (a)
After drawing 1 red: 9 balls left, 3 still blue.
P(2nd blue | 1st red): 39 = 13
P(2nd blue | 1st red) = 13part (b) — both blue
Use multiplication: P(B₁ ∩ B₂) = P(B₁) × P(B₂|B₁)
P(1st blue): 310
P(2nd blue | 1st blue): 2 blue left out of 9 = 29
P(both blue) = 310 × 29 = 690 = 115
P(both blue) = 115
just count what’s left after each draw — that’s all there is to it!
WE 3Conditional probability from a survey
In a class of 30 students, 18 study Spanish, 12 study French, and 5 study both. A student is picked at random.
(a) Given that the student studies Spanish, find the probability they also study French. (b) Given they study French, find the probability they also study Spanish.
Shrunken sample space — narrow down to the “given” group, then count overlap.part (a)
Restrict to Spanish students: 18.
Of those, 5 also study French.
P(French | Spanish) = 518
P(F|S) = 518part (b)
Restrict to French students: 12.
Of those, 5 also study Spanish.
P(Spanish | French) = 512
P(S|F) = 512
P(F|S) and P(S|F) are different — order of conditioning matters!
WE 4Use multiplication to find P(A ∩ B)
For two events A and B: P(A) = 0.4 and P(B|A) = 0.7. Find P(A ∩ B).
P(A) = 0.4, P(B|A) = 0.7
Use the rearranged formula: P(A ∩ B) = P(A) × P(B|A).
Multiply: P(A ∩ B) = 0.4 × 0.7
= 0.28
P(A ∩ B) = 0.28
match the formula to what’s given — A and P(B|A) → use P(A) × P(B|A)
WE 5Use conditional probability to test independence
For two events A and B: P(A) = 0.5, P(B) = 0.4 and P(A|B) = 0.5. Are A and B independent?
Test: does P(A|B) = P(A)?
Compare: P(A|B) = 0.5, P(A) = 0.5
They’re equal!
Yes — A and B are independent (P(A|B) = P(A))
knowing B happened didn’t change the probability of A — that’s independence!
💡 Top tips
- The “given” event goes on the bottom. P(A|B) = P(A ∩ B) ÷ P(B). The bar means “given”, and what’s after the bar is the denominator.
- For “without replacement”, just count what’s left after each draw — usually faster than the formula.
- Three rearrangements come from one formula: P(A ∩ B) = P(B) × P(A|B) = P(A) × P(B|A). Use whichever fits your given values.
- P(A|B) ≠ P(B|A) in general. Order matters!
- To test independence, check P(A|B) = P(A). If they’re equal, independent. If not, they’re not.
- For “given that…” questions, use the shrunken sample space approach: restrict to the “given” group, then count.
- Independence is symmetric: if P(A|B) = P(A), then automatically P(B|A) = P(B) too.
- If P(A|B) > P(A), B happening makes A more likely. If P(A|B) < P(A), B makes A less likely.
⚠ Common mistakes
- Confusing P(A|B) with P(A ∩ B). They’re different! P(A|B) restricts to the B world; P(A ∩ B) is the overall probability of both.
- Putting the wrong event on the bottom of the formula. The “given” event (after the bar) goes in the denominator. P(A|B) divides by P(B), not P(A).
- Assuming P(A|B) = P(B|A). They are usually different — order of conditioning matters.
- Treating “without replacement” problems as independent. They aren’t — the second probability changes after the first draw.
- Forgetting to justify independence answers. You must state the criterion and compare the numbers explicitly.
- Multiplying P(A) × P(B) when events aren’t independent. Only works for independent events; otherwise use the conditional version.
- Mixing up which event “happened first”. P(A|B) means B happened first — even if it’s the second event chronologically.
- Using the formula when you should just count. For frequency tables and simple counts, the shrunken sample space is faster and clearer.
Conditional probability is a key tool — and you’ll see it baked into the next two notes (Venn diagrams and tree diagrams) where it’s the engine that makes everything work.
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