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Cambridge (CIE) IGCSE · International Maths Extended
Tree Diagrams & Multiple Events
The big idea: when one thing happens and then another thing happens, you multiply along the branches — and whenever there are several different ways to get the result you want, you add those possibilities together.
Summary — Everything In This Chapter
Probability tree diagrams map out repeated experiments with two outcomes each, so you can see every possible path an event can take.
Combined probabilities let you calculate "one thing then another" without even drawing a tree — using the golden rule: AND → multiply, OR → add.
Independent events don't affect each other (like two separate dice rolls). Dependent events — especially "without replacement" problems — do.
Conditional probability is the probability of something happening something else has already happened. It's calculated out of a smaller, restricted set of outcomes.
Combined conditional probabilities combine both ideas: successive events where the second one's probability changes because of what happened in the first — most commonly seen in "without replacement" bag/card problems.
1. Probability Tree Diagrams
How do I draw a tree diagram?
Think of a tree diagram as a map of every possible path a two-stage (or more) experiment could take. Each "experiment" — like flipping a coin, drawing a card, or a set of traffic lights — has two outcomes: something happening (A) or it happening (not A).
You read a tree diagram left to right, following the branches. The top-most path all the way across gives you "A B". Every other path down the tree gives you a different combination.
1st experiment 2nd experiment
P(B) ──── B → P(A and B)
P(A) ──── A ┤
┌ └ P(not B) ─ not B → P(A and not B)
START ────┤
└ P(not A) ── not A ┤
├ P(B) ──── B → P(not A and B)
└ P(not B) ─ not B → P(not A and not B)
Every pair of branches coming from the same point must add up to 1. That's because P(not A) = 1 − P(A) — something either happens or it doesn't, there's no third option.
How do I find probabilities from a tree diagram?
Once your tree is drawn and labelled, there are just two moves:
Rule 1 — Multiply along the branchesP(1st outcome 2nd outcome) = P(1st) × P(2nd)Follow one path from left to right, multiplying every probability you pass.
Rule 2 — Add between separate casesP(AA BB) = P(AA) + P(BB)If there's more than one path that satisfies what the question is asking, add those path-probabilities together.
If you're asked for the probability of "at least one" outcome, it's almost always quicker to calculate 1 − P(none) rather than listing and adding every single case that counts as "at least one".
Worked Example — Traffic Lights
A worker drives through two sets of traffic lights. P(1st is green) = 5/7, and P(2nd is green) = 8/9.
Step 3(b) — P(at least one red)?
Method 1 (add the cases): P(G,R) + P(R,G) + P(R,R) = 5/63 + 16/63 + 2/63 = 23/63
Method 2 (the shortcut): 1 − P(G,G) = 1 − 40/63 = 23/63 — same answer, less adding!
Practice Question
A spinner has two independent spins. P(1st spin lands on red) = 3/5 and P(2nd spin lands on red) = 2/3. Find the probability that the spinner lands on red on at least one of the two spins.
Practice Question
Liana has 10 pets: 7 guinea pigs (G) and 3 rabbits (R). She picks one pet at random, takes it away, then picks a second pet at random from the remaining 9. Find P(she picks two different kinds of animal).
2. Combined Probabilities
The AND / OR rule
You don't always need to draw a full tree. If you can describe what you want using the words "and" / "or", you can calculate it directly. This is exactly the same maths as the tree diagram — just without drawing the picture.
The Golden Rule"AND" means multiply. "OR" means add.P(A and B) = P(A) × P(B) | P(AA or BB) = P(AA) + P(BB)
The real skill here is translating the question into "and/or" language first. For example, flipping a coin twice:
P(two heads) = P(head and head)
P(both the same) = P(head and head) or (tail and tail) = P(HH) + P(TT)
What does "independent" mean?
Independent events are events where the outcome of one has zero effect on the probability of the other. Rolling a die and flipping a coin are independent — the die doesn't "know" what the coin did.
"Without replacement" is NOT independent. If you take a red card out of a deck and don't put it back, you've changed the deck for the next draw — there's one fewer card, and one fewer red card. This is one of the most common places students lose marks, because it similar to independent events but isn't.
Worked Example — Counters With Replacement
A box has 3 blue and 8 red counters (11 total). A counter is drawn, its colour noted, then put back. A second counter is drawn.
(b) P(different colours) = P(B and R) + P(R and B) = (3/11 × 8/11) + (8/11 × 3/11) = 24/121 + 24/121 = 48/121
Practice Question
A fair coin is flipped 3 times. Using the AND/OR rule (not a full tree), find the probability of getting exactly two heads and one tail, in any order.
3. Conditional Probability
What is a conditional probability?
A conditional probability is the probability of something happening (A), given that something else (B) has happened. The key shift in thinking: you're no longer working out of the whole sample space — you're working out of a smaller, restricted set of outcomes, because B has already narrowed things down.
Notation (recognise it — you won't need to reproduce it)P(A | B) = "the probability of A, given that B has already happened"You're expected to calculate conditional probabilities using Venn diagrams, tree diagrams, or tables — not necessarily using this notation directly.
Here's the clearest way to see the difference. Imagine a computer randomly picks a digit from {1,2,3,4,5,6,7,8,9}:
Not conditional: P(multiple of 3) = 3/9 (three numbers — 3, 6, 9 — out of )
Conditional: P(multiple of 3 it's even) = 1/4 — because if we it's even, we're only choosing from {2,4,6,8}, and only 6 is a multiple of 3.
Conditional probability questions almost always contain the phrase "given that...". Whenever you see it, immediately ask yourself:
Worked Example — Venn Diagram
A = {2, 6, 12, 14, 28}, B = {7, 14, 21, 28, 35}. A number is picked at random. Find P(number is in A it is in B).
Step 1 — Since we're told it's in B, our restricted pool is just the 5 numbers in B: {7, 14, 21, 28, 35}.
Step 2 — Of those 5 numbers, which are also in A? Just 14 and 28 — that's 2 numbers.
Answer: P(A | B) = 2/5
Practice Question
In a class, 18 students study French, 12 study Spanish, and 5 study both. A student is picked at random from those who study French. Find the probability they also study Spanish.
4. Combined Conditional Probabilities
Two events in a row, where the second depends on the first
This is where everything from this chapter comes together. You have two (or more) successive events — and the second event's probability is conditional on (changed by) what happened in the first. The classic example is drawing cards or counters without replacement.
The adjustment ruleP(red then red, no replacement) = 26/52 × 25/51Notice: the denominator drops by 1 each time (fewer total items left), and the numerator drops too if you removed one of "that" type.
If a question says "two cards are drawn" (with no mention of replacing), you should assume it means one card is drawn, then a second is drawn without replacement. The maths works out identically either way you picture it.
What if there are multiple possible orders?
Sometimes a question wants "one of each" or "all the same" — which can happen in several different orders. You need a listing strategy so you don't miss any, then add every case together.
Remember: AB and BA are different — AB means A happened first, then B; BA means B happened first, then A.
List every valid order (e.g. AAB, ABA, BAA), calculate each one's probability, then add them all with the "or" rule.
Try not to simplify your fractions too early. Keeping the same denominator across every case (e.g. everything over /380) makes adding them together at the end far easier and less error-prone.
Worked Example — Beads Without Replacement
A bag has 10 yellow (Y), 6 blue (B), 4 green (G) beads — 20 total. One bead is drawn and not replaced, then a second is drawn.
Find P(both beads are the same colour) — three possible cases: YY, BB, GG.
Worked Example — Using Algebra (Harder / Exam-style)
A bag has only red and blue marbles. There are 3 more red than blue. Kai takes 2 marbles at random, and P(both blue) = 1/7. Find how many blue marbles there were initially.
Step 1 — Set up expressions. Let x = number of blue marbles, so red = x + 3, and total = 2x + 3.
Step 2 — P(1st is blue) = x / (2x + 3)
Step 3 — P(2nd is blue | 1st was blue) = (x − 1) / (2x + 2)
Step 4 — Multiply and set equal to 1/7:
[x/(2x+3)] × [(x−1)/(2x+2)] = 1/7
Step 6 — Factorise: (3x + 1)(x − 6) = 0 → x = −1/3 or x = 6
Answer: x must be a positive whole number, so x = 6 blue marbles.
Practice Question
A drawer has 5 black socks and 4 white socks. Two socks are taken out at random, one after another, without replacement. Find the probability that both socks are the same colour.
Practice Question
Using the same drawer (5 black, 4 white socks, no replacement), find the probability that the two socks are different colours, by listing the possible orders.
What to Memorise
Term / Rule
What it means
P(not A) = 1 − P(A)
Branches from the same point always sum to 1.
AND → multiply
P(A and B) = P(A) × P(B). Follow one path along a tree.
OR → add
P(A or B) = P(A) + P(B), when A and B can't both be the exact same single outcome you're combining.
"At least one" shortcut
P(at least one) = 1 − P(none). Almost always faster than listing every case.
Independent events
One event has no effect on the other's probability (e.g. "with replacement").
Without replacement
NOT independent — the denominator (and sometimes numerator) shrinks by 1 for the next draw.
Conditional probability, P(A|B)
Probability of A, given B already happened — calculated from a smaller restricted pool of outcomes, not the full total.
"Given that..."
The exam's signal phrase for a conditional probability question.
Order matters in listing
AB ≠ BA. List every valid order separately before adding.
Concepts Checklist
Exam Tips & Common Mistakes
Don't cancel fractions too early
Keep fractions like 5/63 rather than simplifying mid-calculation. It's far easier to add fractions together when they already share the same denominator — simplify only at the very end.
"Without replacement" ≠ independent
This is the single most common error in this topic. If an item isn't put back, the second draw's probabilities MUST change — both the numerator (if relevant) and the denominator.
Conditional ≠ "out of everything"
When you see "given that", resist the urge to just divide by the total sample space. You must divide by the size of the restricted group named in the "given that" clause.
AB and BA are different outcomes
When listing orders for "one of each" or similar questions, always check whether you need to count both orderings separately — a very common place marks are dropped.
Use the complement for "at least one"
Examiners love "at least one" questions specifically because students try to list every case. Train yourself to immediately reach for 1 − P(none) instead.
Sanity-check your tree
Once you've calculated all the end-of-branch probabilities in a tree, they should always sum to exactly 1. This is a free way to catch arithmetic mistakes before submitting an answer.
Show every step of algebraic setups
For "find x" style conditional probability questions, examiners award method marks for correctly setting up the fractions before you even solve the equation — so write out P(1st) and P(2nd | 1st) explicitly.