OCR section 2.03 builds from the addition and multiplication rules to the A-level headline: conditional probability. You will use Venn diagrams, tree diagrams and two-way tables, and learn to test independence properly.
Draw the diagram first. Always. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.
Are they independent? P(A) × P(B) = 0.4 × 0.5 = 0.2, which equals P(A ∩ B). So yes, A and B are independent.
Are they mutually exclusive? No — P(A ∩ B) = 0.2 ≠ 0.
Do not confuse the two. Mutually exclusive and independent are almost opposites: if A and B are mutually exclusive and both have non-zero probability, then knowing A happened tells you B definitely did not — which is the strongest possible dependence.
P(A | B) reads ‘the probability of A given that B has happened’. Knowing B restricts you to the part of the sample space where B is true — so you rescale.
P(A | B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4.
Notice P(A | B) = 0.4 = P(A). Knowing that B happened changed nothing — which is precisely what independence means. In fact:
The classic error: P(A | B) is not the same as P(B | A). ‘The probability it is raining given the ground is wet’ is very different from ‘the probability the ground is wet given it is raining’.
A Venn diagram lets you fill in the overlap first and work outwards — the fastest way to avoid double counting.
Of 30 students, 18 study Maths, 14 study Physics, and 6 study both.
Maths only = 18 − 6 = 12. Physics only = 14 − 6 = 8. Both = 6.
Total inside the circles = 12 + 8 + 6 = 26.
∴ students studying neither = 30 − 26 = 4.
Cross-check with the addition rule: P(M ∪ P) = 18/30 + 14/30 − 6/30 = 26/30. Neither = 1 − 26/30 = 4/30. ✓
Notation: A′ means not A (the complement), and P(A′) = 1 − P(A). ‘Neither’ is the region (A ∪ B)′.
On a tree diagram you multiply along the branches and add between different paths that give the same outcome.
There are 8 counters to start; after one is taken, only 7 remain — so the second branch probabilities change. That is conditional probability in action.
P(both red) = (5/8) × (4/7) = 20/56 = 0.357 (3 s.f.).
P(exactly one red) = P(red then blue) + P(blue then red)
= (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56 = 0.536 (3 s.f.).
Check the total: P(both blue) = (3/8)(2/7) = 6/56. And 20/56 + 30/56 + 6/56 = 56/56 = 1. ✓
With replacement, the counter goes back and the denominators stay at 8 — the draws are then independent. Without replacement they are not. Read the question carefully; this single word changes every branch.
Tap a scenario, then tap the group it belongs to.
Tap a description on the left, then the rule it corresponds to.
Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) — subtract the overlap
Mutually exclusive: P(A ∩ B) = 0
Independent: P(A ∩ B) = P(A) × P(B), equivalently P(A | B) = P(A)
Conditional: P(A | B) = P(A ∩ B) ÷ P(B) — and P(A | B) ≠ P(B | A)
Venn: fill the overlap first; P(A′) = 1 − P(A)
Trees: multiply along branches, add between paths; without replacement changes the second row
Check: all the mutually exclusive outcomes must total exactly 1
That is OCR 2.03 — and conditional probability is the engine behind the distributions topic. Press Finish to see your score.
You've worked through Probability for OCR A-level Mathematics A. 🎉
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