Probability Tree Diagrams – Step by Step Guide

Why Tree Diagrams Matter

Probability tree diagrams are one of the most useful tools in IGCSE Maths. They turn complicated probability questions — the kind that are hard to think through in your head — into a systematic visual process. If you can draw a correct tree diagram, you can answer almost any probability question the exam throws at you.

This guide takes you through the process step by step, from the simplest cases to the trickier questions involving dependent events (without replacement).

The Two Golden Rules

Every tree diagram question relies on just two rules:

Rule 1: Multiply along branches to find the probability of a specific sequence of events happening. This is the AND rule — event A AND event B.

Rule 2: Add between branches to find the probability of one outcome OR another. This is the OR rule.

If you remember nothing else, remember: multiply along, add between.

How to Draw a Tree Diagram

Step 1: Identify the Events

Each stage of the experiment (each event) gets its own set of branches. If a coin is flipped twice, there are two sets of branches. If a ball is drawn from a bag twice, there are two sets of branches.

Step 2: Draw the Branches

From each point, draw branches for every possible outcome. Label each branch with the outcome and its probability. The probabilities on branches from the same point must add up to 1.

Step 3: Calculate End Probabilities

At the end of each complete path through the tree, multiply all the probabilities along that path.

Worked Example 1: Independent Events

Question: A bag contains 3 red balls and 5 blue balls. A ball is drawn, its colour recorded, and it is put back. A second ball is then drawn. Find the probability of getting (a) two red balls, (b) exactly one red ball.

Setting up the tree:

First draw:

• Red: P = 3/8
• Blue: P = 5/8

Second draw (since the ball is replaced, probabilities are the same):

• Red: P = 3/8
• Blue: P = 5/8

The four possible paths:

Path	Calculation	Probability
Red, Red	3/8 × 3/8	9/64
Red, Blue	3/8 × 5/8	15/64
Blue, Red	5/8 × 3/8	15/64
Blue, Blue	5/8 × 5/8	25/64

Check: 9/64 + 15/64 + 15/64 + 25/64 = 64/64 = 1. Correct.

(a) P(two red) = 9/64 (multiply along the Red, Red path)

(b) P(exactly one red) = 15/64 + 15/64 = 30/64 = 15/32 (add the two paths that contain exactly one red)

Worked Example 2: Dependent Events (Without Replacement)

Question: A bag contains 4 green balls and 6 yellow balls. A ball is drawn and NOT replaced. A second ball is then drawn. Find the probability of getting (a) two green balls, (b) one of each colour.

Setting up the tree:

This is where it gets more interesting. Because the first ball is not replaced, the probabilities change for the second draw.

First draw (10 balls total):

• Green: P = 4/10
• Yellow: P = 6/10

Second draw after Green first (9 balls left, 3 green, 6 yellow):

• Green: P = 3/9
• Yellow: P = 6/9

Second draw after Yellow first (9 balls left, 4 green, 5 yellow):

• Green: P = 4/9
• Yellow: P = 5/9

The four paths:

Path	Calculation	Probability
Green, Green	4/10 × 3/9	12/90
Green, Yellow	4/10 × 6/9	24/90
Yellow, Green	6/10 × 4/9	24/90
Yellow, Yellow	6/10 × 5/9	30/90

Check: 12/90 + 24/90 + 24/90 + 30/90 = 90/90 = 1. Correct.

(a) P(two green) = 12/90 = 2/15

(b) P(one of each) = 24/90 + 24/90 = 48/90 = 8/15

Exam tip: Always simplify your fractions in the final answer. The mark scheme typically shows the simplified form. Also, always check that your end probabilities sum to 1 — if they do not, you have made an error somewhere.

Worked Example 3: Three Events

Question: The probability that it rains on any day is 0.3. Find the probability that in three consecutive days, it rains on exactly two of the three days.

Setting up the tree:

Each day has two outcomes: Rain (0.3) and No Rain (0.7). With three days, the tree has three stages and eight end paths.

Rather than drawing all eight paths, we can identify just the paths with exactly two rainy days:

Path	Calculation	Probability
Rain, Rain, No Rain	0.3 × 0.3 × 0.7	0.063
Rain, No Rain, Rain	0.3 × 0.7 × 0.3	0.063
No Rain, Rain, Rain	0.7 × 0.3 × 0.3	0.063

P(exactly 2 rainy days) = 0.063 + 0.063 + 0.063 = 0.189

Exam tip: With three-stage tree diagrams, it is easy to miss a path. Count systematically. For “exactly 2 out of 3,” there are always 3 paths (the non-event can occur on day 1, day 2, or day 3).

Worked Example 4: Conditional Probability Setup

Question: In a class, 60% of students are girls. 80% of girls pass a test and 70% of boys pass the test. A student is chosen at random. Find the probability that the student passed the test.

Setting up the tree:

First branch — gender:

• Girl: P = 0.6
• Boy: P = 0.4

Second branch — test result:

• Girl who passes: P = 0.8
• Girl who fails: P = 0.2
• Boy who passes: P = 0.7
• Boy who fails: P = 0.3

Paths where the student passes:

Path	Calculation	Probability
Girl, Pass	0.6 × 0.8	0.48
Boy, Pass	0.4 × 0.7	0.28

P(passes) = 0.48 + 0.28 = 0.76

Common Mistakes to Avoid

1. Forgetting to change probabilities for “without replacement” questions. If the question says the item is not replaced, the total number and the relevant count both change for the second draw.
2. Missing paths. When asked for “at least one” or “exactly one,” make sure you identify all relevant paths. It often helps to list all paths systematically first.
3. Adding when you should multiply (or vice versa). Remember: along a single path you multiply; across different paths you add.
4. Not simplifying the final answer. Always reduce fractions to their simplest form.
5. Using “at least one” the hard way. For “at least one” questions, it is often easier to calculate P(none) and subtract from 1: P(at least one) = 1 - P(none).

Exam Tips

• Draw your tree diagram clearly with enough space between branches. A cramped diagram leads to errors.
• Label every branch with both the outcome and its probability.
• Check that probabilities from each point add up to 1.
• Check that all end-path probabilities add up to 1.
• Show your multiplication along branches and your addition between branches clearly — both earn method marks.

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