Probability

Sample spaces, Venn diagrams, tree diagrams & the addition rule. ✎

① The Basics

P(A) = number of favourable outcomes total number of outcomes

0 ≤ P(A) ≤ 1 always. 0 = impossible, 1 = certain.

Complement: P(A′) = 1 − P(A) ("not A")

All the probabilities of a full sample space add to 1.

"At least one" is nearly always easier as 1 − P(none).

② Venn Diagrams

intersection — "A and B", the overlap.

union — "A or B", everything in either circle.

complement — everything outside.

Always fill the overlap first, then work outwards so each region isn't double-counted.

③ Addition Rule ✦ formula booklet

P(A∪B) = P(A) + P(B) − P(A∩B)

You subtract the overlap because adding both circles counts it twice.

Mutually exclusive

P(A∪B) = P(A) + P(B)

④ Tree Diagrams

Multiply ALONG the branches (that's "and").

Add DOWN the final outcomes you want (that's "or").

Each pair of branches must sum to 1 — a fast self-check.

All the end results also add to 1 (0.42+0.18+0.20+0.20 = 1 ✓).

★ Example — without replacement ✎

A bag has 5 red and 3 blue counters. Two are drawn without replacement. Find P(both red) and P(one of each).

First draw: P(red) = 5/8. Then only 7 left, 4 of them red.

P(both red)

One of each

(5/8)(3/7) = 15/56

(3/8)(5/7) = 15/56

Add them: 15/56 + 15/56 = 30/56 = 15/28

Tip!

"One of each" always needs both orders. Forgetting the second route halves your answer — the single most common probability slip.

IB MATHEMATICS · ANALYSIS & APPROACHES · SL · TOPIC 4.6

Conditional Probability

"Given that…", independence & reading probabilities from a two-way table. ✎

① The Conditional Formula ✦ formula booklet

P(A|B) reads "the probability of A given that B has happened".

P(A|B) = P(A∩B) / P(B)

The condition shrinks the sample space — B becomes your new "whole world", which is why you divide by P(B).

② Independence

Independent = one event doesn't change the other's probability.

P(A∩B) = P(A) × P(B)

Equivalently: P(A|B) = P(A) — knowing B tells you nothing.

To prove independence: compute both sides and show they're equal.

With replacement → independent. Without → not independent.

③ Two-Way Tables

The fastest way to read conditional probabilities — the condition picks the row or column total.

P(Bus) = 33/50 ← whole table

P(Bus | Girl) = 18/30 = 0.6 ← girls row only

P(Girl | Bus) = 18/33 ≈ 0.545 ← bus column only

Notice P(A|B) ≠ P(B|A) — same overlap, different denominator.

★ Example — using the formula ✎

P(A) = 0.6, P(B) = 0.5, P(A∪B) = 0.8. Find P(A∩B), then P(A|B). Are A and B independent?

Rearrange the addition rule: 0.8 = 0.6 + 0.5 − P(A∩B) → P(A∩B) = 0.3

Independence test: P(A) × P(B) = 0.6 × 0.5 = 0.3 = P(A∩B) ✓

Yes, independent — and notice P(A|B) = 0.6 = P(A), the same conclusion.

The addition rule is your way in whenever P(A∩B) is missing. Then show the comparison for independence — don't just assert it.

★★ Example — conditional on a tree ✎

A factory has two machines. Machine A makes 60% of items, 5% faulty; Machine B makes 40%, 8% faulty. An item is faulty — what is the probability it came from A?

Multiply along each faulty route:

(the denominator)

P(A | faulty)

Sensible: B is more likely to be faulty, so less than half of faulty items come from A.

Working backwards along a tree: the denominator is the total of every route that gives the condition. Add all faulty paths, then take the one you want.