IB Maths AI SL Probability Paper 1 & 2 Basics ~7 min read

Probability & Types of Events

Probability is the number of favourable outcomes divided by the total number of possible outcomes. This note builds the vocabulary: sample space, event, complement, intersection, union, conditional. Once you know what each symbol means, every probability question becomes a matter of identifying which formula applies and counting carefully.

📘 What you need to know

Basic formula: P(A) = n(A)n(U) — favourable outcomes over total outcomes (when all outcomes are equally likely).
Complement: P(A‘) = 1 − P(A). The probability that A does NOT happen.
Intersection A ∩ B: both happen (“A AND B”). Union A ∪ B: at least one happens (“A OR B or both”).
Union formula: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Subtract the overlap so it isn’t counted twice.
Expected occurrences: in n trials with probability p per trial, expect roughly np successes. Not exact — an average.
Theoretical vs experimental: theoretical = use the formula; experimental (relative frequency) = use observed counts ÷ trials.

The vocabulary — sample space and events

An experiment is a repeatable activity. A trial is one repetition. An outcome is one possible result. An event is a collection of outcomes you care about. The sample space U is the set of ALL possible outcomes — the universe of the experiment.

For one roll of a fair six-sided die: U = {1, 2, 3, 4, 5, 6}; an event like “even number” is the subset {2, 4, 6} containing 3 outcomes. So P(even) = 3/6 = 1/2.

Combined events: AND, OR, NOT

Two events A and B in the same sample space create four regions: outcomes in A only, in B only, in both, or in neither. The Venn diagram makes this concrete:

The Venn diagram organises every outcome into one of four regions. Intersection = the overlap (both events); Union = anything inside either circle (at least one event); Complement of A = everything outside the A circle.

Union formula: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The subtraction is essential — without it, the overlap would be counted twice (once in P(A) and again in P(B)).

Core formulas (in the booklet) P(A) = n(A)n(U) P(A‘) = 1 − P(A) P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Expected occurrences and experimental probability

Expected occurrences: if an event has probability p and you run n trials, you expect about np successes. Flip a fair coin 100 times: expect 50 heads (you won’t get exactly 50 every time, but on average across many runs, that’s the figure).

Experimental probability (a.k.a. relative frequency): when you’ve actually run trials, divide the observed count by the total number of trials. With enough trials this converges toward the theoretical probability.

Theoretical vs experimental: theoretical uses the SAMPLE SPACE; experimental uses OBSERVED COUNTS. They should be similar if the experiment is fair and you’ve run enough trials.

🧭 Recipe — any basic probability question

Identify the sample space: list outcomes or use a table for two-stage experiments. Count n(U).
Identify the event(s): list (or count) the outcomes that match the event. Call this n(A).
Apply the basic formula P(A) = n(A) / n(U). Simplify to lowest terms when expressing as a fraction.
For combined events, pick the right formula: complement (1 − P), union (add and subtract overlap), or intersection (multiply — covered in the next note).
Check the answer makes sense: 0 ≤ P ≤ 1; expected values shouldn’t exceed the number of trials.

Worked examples

WE 1

List a sample space and compute a probability

Two fair six-sided dice are rolled and their scores added. (a) State the size of the sample space. (b) Find the probability that the sum is at least 9.

(a) sample space = all ordered pairs (a, b) 6 × 6 = 36 outcomes (b) list pairs with a + b ≥ 9 sum = 9: (3,6), (4,5), (5,4), (6,3) — 4 pairs sum = 10: (4,6), (5,5), (6,4) — 3 pairs sum = 11: (5,6), (6,5) — 2 pairs sum = 12: (6,6) — 1 pair total = 4 + 3 + 2 + 1 = 10 apply formula P = 10/36 = 5/18 (a) 36 outcomes · (b) P = 5/18 always simplify fractions: 10/36 = 5/18 by dividing top and bottom by 2. A systematic 6×6 table is the safest way to count without missing pairs.

WE 2

Expected number of occurrences

A biased coin lands on heads with probability 0.6. The coin is flipped 250 times. Find the expected number of heads.

use expected value formula expected = n × p = 250 × 0.6 = 150 heads 150 heads “expected” means the AVERAGE you’d get if you repeated the 250 flips many times. A single run might give 142 or 156 — that’s normal variation, not a contradiction.

WE 3

Complement — use it as a shortcut

The probability that it rains on a given day in November is 7/20. (a) Find the probability that it does NOT rain on a given day. (b) Find the probability that it rains on at least one day in a 2-day window, assuming the days are independent.

(a) complement P(no rain) = 1 − 7/20 = 20/20 − 7/20 = 13/20 (b) “at least one rainy day” opposite event = “no rain on both days” P(no rain both) = (13/20)² = 169/400 P(at least one rainy day) = 1 − 169/400 = 231/400 ≈ 0.578 (a) 13/20 · (b) 231/400 ≈ 0.578 “at least one” almost always means use the complement: P(at least one) = 1 − P(none). Direct counting of “exactly 1 OR exactly 2” works but is slower.

WE 4

Union formula in action

In a music school, P(plays piano) = 0.55, P(plays guitar) = 0.4, and P(plays both) = 0.2. (a) Find the probability that a randomly chosen student plays at least one of the two instruments. (b) Find the probability the student plays neither.

let P = piano, G = guitar (a) “at least one” = P(P ∪ G) P(P ∪ G) = P(P) + P(G) − P(P ∩ G) = 0.55 + 0.4 − 0.2 = 0.75 (b) “plays neither” = complement of (a) P(neither) = 1 − 0.75 = 0.25 (a) 0.75 · (b) 0.25 always SUBTRACT the overlap when adding two probabilities. Without the − 0.2 step, the answer would be 0.95 — an over-count, because the 20% who play both are in both P(P) and P(G).

WE 5

Without replacement — conditional in disguise

A box contains 12 cards: 5 red and 7 blue. Two cards are drawn at random without replacement. Find the probability that both are red.

P(both red) = P(1st red) × P(2nd red | 1st red) Step 1 — first card P(1st red) = 5/12 Step 2 — second card (given 1st was red) now 11 cards left, 4 red P(2nd red | 1st red) = 4/11 multiply P(both red) = (5/12) × (4/11) = 20/132 = 5/33 ≈ 0.152 P(both red) = 5/33 ≈ 0.152 key feature of “without replacement”: the denominator DROPS by 1 for the second pick (12 → 11), and the count of the chosen colour drops by 1 too (5 → 4). For “with replacement”, both stay the same.

WE 6

Experimental probability and prediction

A four-colour spinner is spun 200 times. The results: red 48, blue 62, green 53, yellow 37. (a) Find the experimental probability of landing on blue. (b) Use this to estimate the expected number of blue results in the next 500 spins.

(a) relative frequency formula P_exp(blue) = frequency of blue / total = 62 / 200 = 31/100 = 0.31 (b) use np with this experimental probability expected blue = 500 × 0.31 = 155 (a) P_exp(blue) = 0.31 · (b) ~155 blue results if a fair spinner gave perfect theoretical 0.25 per colour, you’d expect 125 blue. The experimental 0.31 suggests the spinner might be slightly biased toward blue — or it’s just sample noise. Either way, the BEST estimate from this data is 0.31.

💡 Top tips

“At least one” = use the complement: 1 − P(none) is almost always faster than counting cases.
Draw a 2-way table or Venn whenever there are two events — it organises the four regions automatically.
Simplify fractions: write 5/18, not 10/36. Most exam mark schemes accept either, but simplified is safer.
Probabilities live in [0, 1]: if you get P > 1 or P < 0, you’ve made an arithmetic error.
Expected = np isn’t an exact count — phrase the answer as “approximately” or “expect about” in context.

⚠ Common mistakes

Adding probabilities without subtracting the overlap: P(A ∪ B) = P(A) + P(B) ONLY if the events can’t happen together (mutually exclusive). Otherwise subtract P(A ∩ B).
Treating “with replacement” and “without replacement” the same: the second-pick denominator changes for without-replacement problems.
Confusing complement with mutually exclusive: complement = “A doesn’t happen”; mutually exclusive = “A and B can’t both happen”. Not the same idea.
Forgetting n(U) is the DENOMINATOR: writing 5/2 instead of 2/5 is a costly slip.
Saying “expected 50 heads means I’ll get exactly 50”: it’s an average over many runs, not a single-run prediction.

Next up: Independent and Mutually Exclusive Events. Two special cases of combined events with their own simplified rules — for independent events P(A ∩ B) = P(A)P(B), and for mutually exclusive events P(A ∩ B) = 0, so the union formula simplifies to P(A) + P(B). Knowing which case applies is half the battle.

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