Key Idea: Probability measures how likely an event is, on a scale from 0 to 1. It runs through both papers — single events, 'at least one' questions, draws without replacement, and how many times to expect something over many trials.
🎲 Probability of a single event
- number of favourable outcomes
- total number of equally likely outcomes
0 ≤ P(A) ≤ 1 always. P = 0 means impossible, P = 1 means certain. If you ever get a value above 1 or below 0, you've miscounted — go back and recheck.
🧩 The four core moves
| Situation | What to do | Quick example |
|---|---|---|
| Single event | Count favourable ÷ total. | 5 reds of 10 → P = 5/10 = 1/2 |
| Complement / 'at least one' | P(A′) = 1 − P(A); P(at least one) = 1 − P(none). | P(at least one six) = 1 − P(no six) |
| Two dice (sample space) | Grid of 6 × 6 = 36 ordered outcomes; count matches. | Total = 7 → 6 cells → 6/36 |
| Without replacement | Multiply along the chain, reduce both totals each draw. | 4/10 × 3/9 = 2/15 |
🔁 Expected number = n × P
- number of times the trial is repeated
- probability of the event on one trial
Tip: An expected number is a long-run average, not a single count — so 7.5 is a perfectly valid answer. If P isn't given, work it out first, then multiply by n.
✏️ IB-style worked examples
IB-style question — single event and its complement
A drawer holds 7 black and 3 grey socks. One sock is taken at random. (a) Find the probability it is grey. (b) Find the probability it is not grey.
Step by step:
Total socks, then favourable ÷ total for grey.
'Not grey' is the complement: 1 − P(grey).
P(grey) = 3/10; P(not grey) = 7/10.
IB-style question — sample space and 'at least one'
Two fair dice are rolled. (a) Find the probability the two scores are equal. (b) Find the probability of getting at least one five.
Step by step:
Grid has 6 × 6 = 36 outcomes. Equal pairs: (1,1)…(6,6) — 6 of them.
For 'at least one five', flip to P(no five on either die).
Take the complement.
P(equal) = 1/6; P(at least one five) = 11/36.
IB-style question — drawing without replacement
A jar contains 5 lime and 4 cherry sweets. Two are eaten, one after the other. Find the probability both are lime.
Step by step:
First lime: 5 of the 9 sweets.
Now 4 lime of 8 left. Multiply along the chain.
P(both lime) = 5/18.
IB-style question — expected number over many trials
A spinner has 10 equal sectors, 4 of them yellow. The spinner is spun 150 times. Find the expected number of yellows.
Step by step:
Probability of yellow on one spin.
Expected number = n × P.
60 yellows expected.
Important: Without replacement, the counts change after the first draw: there is one fewer item and one fewer of that type. So the second probability has totals like 3/9, not 3/10 — and reduce both numbers, not just the top.
Tap each card to reveal the answer.
P(A) = 0.35. Find P(A′). 0.65 — the complement is 1 − 0.35.
Two dice: P(total = 7)? 6/36 = 1/6 — six ordered cells give a total of 7.
Fast way to do 'at least one'? Use 1 − P(none) instead of counting every case.
Box: 3 red, 5 blue. Draw 2 (no replacement). P(both red)? 3/8 × 2/7 = 3/28 — reduce both totals on the second draw.
P = 0.2, repeated 80 times. Expected number? 80 × 0.2 = 16 occurrences expected.
Can an expected number be 4.5? Yes — it's a long-run average, so decimals are fine.
Exam Tips
- Every probability is between 0 and 1 — a value outside that range means a counting error.
- P(A′) = 1 − P(A); for 'at least one', use 1 − P(none).
- Two dice → 36 ordered outcomes; (2,5) and (5,2) are different cells.
- Without replacement: reduce both totals each draw (e.g. 4/10 then 3/9).
- Expected number = n × P — show P and the multiplication; a decimal answer is fine.