Tree diagrams | OCR GCSE Maths Notes

Visual model

Multiply along branches, add separate routes

A

A'

second eventone complete path = one outcome

Multiply along one path.

Add separate paths.

Update probabilities when items are not replaced.

Gold-standard guide

24 mins

What you will learn

I can draw a tree diagram with correct probabilities summing to 1 at each split.

I can find P(combined outcome) by multiplying along a route.

I can find P(event) by adding all qualifying routes.

I can update branch probabilities for without-replacement problems.

I can use 1 − P(none) to solve 'at least one' problems efficiently.

Useful before you start

Fractions, decimals and percentagesMultiplying fractionsAdding fractions and decimalsComplement rule: P(A) + P(not A) = 1Basic probability (sample space, outcomes)

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Trial

One action in a probability experiment — e.g. one coin flip or one draw from a bag.

Step 1

Branch

An arrow on a tree diagram showing one possible outcome and its probability.

Step 2

Route (path)

A sequence of branches from start to finish — multiply along it to get the route probability.

Step 3

Independent events

Two events where the first outcome does not affect the probability of the second.

Step 4

Without replacement

An item drawn is not returned, so the total and counts change for the next draw.

Step 5

Complementary probability

P(event) = 1 − P(event doesn't happen) — fastest for 'at least one' questions.

Multiply along one route

P(outcome A and then outcome B) = P(A) × P(B on second branch, which may depend on A).

Add separate routes

P(event E) = sum of probabilities of all routes in which event E occurs. Only add probabilities of routes that qualify.

Complement rule

P(at least one success) = 1 − P(no successes). Often faster than listing all 'at least one' routes.

Without replacement adjustment

After removing one item, reduce the total by 1 and, if a specific type was removed, reduce that type's count by 1 too.Update every second-trial branch accordingly.

Worked example

A bag contains 4 red counters and 3 blue counters. Two counters are drawn without replacement. Find P(one of each colour).

There are two successful routes: Red then Blue (RB) and Blue then Red (BR).

P(RB) = P(R first) × P(B second | R first) = $\frac{4}{7}$ × $\frac{3}{6}$ = $1\frac{2}{4}2$ .

P(BR) = P(B first) × P(R second | B first) = $\frac{3}{7}$ × $\frac{4}{6}$ = $1\frac{2}{4}2$ .

P(one of each) = $1\frac{2}{4}2$ + $1\frac{2}{4}2$ = $2\frac{4}{4}2$ = $\frac{4}{7}$ .

Final answer

$\frac{4}{7}$

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Diagnosticrecall

The probability that it rains on any day is 0.3. What is the probability that it does not rain?

1 mark1 mintree-d1

Show solution

Worked solution

1.All probabilities for the outcomes of one trial must sum to 1.
2.P(no rain) = 1 − P(rain) = 1 − 0.3.

Final answer

0.7

Mark points

B1: 0.7 (accept $\frac{7}{1}0$ or any exact equivalent).

Watch out

Subtracting from 100 (giving 97 or 70) because the student is thinking of percentages without converting, or subtracting from 10 and getting 9.7.

Easyprocedure

A biased coin has P(Heads) = 0.6. The coin is flipped twice. Find P(two Heads).

2 marks2 minstree-e1

Show solution

Worked solution

1.The flips are independent: the second probability is unaffected by the first.
2.Multiply along the HH route: P(HH) = 0.6 × 0.6.
3.P(HH) = 0.36.

Final answer

0.36

Mark points

M1: recognises multiplication is needed: 0.6 × 0.6 seen.
A1: 0.36.

Watch out

Adding the probabilities along the route (0.6 + 0.6 = 1.2), or using P(H) = 0.6 only once instead of applying it twice.

Mediumreasoning

A spinner has P(red) = 0.4 and P(blue) = 0.6. It is spun twice. Find P(exactly one red).

3 marks3 minstree-m1

Show solution

Worked solution

1.Two routes give exactly one red: Red then Blue (RB) and Blue then Red (BR).
2.P(RB) = 0.4 × 0.6 = 0.24.
3.P(BR) = 0.6 × 0.4 = 0.24.
4.P(exactly one red) = 0.24 + 0.24 = 0.48.

Final answer

0.48

Mark points

M1: correctly identifies both routes RB and BR (award for calculating either route correctly).
M1: correctly calculates probability of one route as 0.24.
A1: 0.48 (from correctly adding both routes).

Watch out

Counting only one of the two successful routes (finding only RB or only BR but not both), giving 0.24 instead of 0.48.

Hardreasoning

A bag has 5 red and 3 blue counters. Two are drawn without replacement. Find P(at least one blue counter).

4 marks4 minstree-h1

Show solution

Worked solution

1.Use the complement: P(at least one blue) = 1 − P(no blue) = 1 − P(both red).
2.P(first red) = $\frac{5}{8}$ .
3.P(second red | first red) = $\frac{4}{7}$ (only 4 red remain from 7 counters).
4.P(both red) = $\frac{5}{8}$ × $\frac{4}{7}$ = $2\frac{0}{5}6$ = $\frac{5}{1}4$ .
5.P(at least one blue) = 1 − $\frac{5}{1}4$ = $\frac{9}{1}4$ .

Final answer

$\frac{9}{1}4$

Mark points

M1: recognises complement strategy: 1 − P(both red) set up.
M1: correct without-replacement probabilities $\frac{5}{8}$ and $\frac{4}{7}$ both seen.
M1: correctly computes $\frac{5}{8}$ × $\frac{4}{7}$ = $\frac{5}{1}4$ .
A1: 1 − $\frac{5}{1}4$ = $\frac{9}{1}4$ .

Watch out

Attempting to add all 'at least one blue' routes individually (BB, RB, BR) but using with-replacement probabilities of $\frac{3}{8}$ on every second branch instead of updating the counts.

Very Hardproblem solving

In a box there are 6 chocolates: 4 dark and 2 milk. Three chocolates are chosen at random without replacement. Find the probability that the three chocolates include more dark than milk.

5 marks5 minstree-vh1

Show solution

Worked solution

1.The event 'more dark than milk' in three chocolates means: exactly 2 dark and 1 milk, OR exactly 3 dark and 0 milk.
2.P(3 dark) = $\frac{4}{6}$ × $\frac{3}{5}$ × $\frac{2}{4}$ = $2\frac{4}{1}20$ = $\frac{1}{5}$ .
3.P(2 dark, 1 milk): the milk chocolate can be in position 1, 2 or 3.
4.P(milk in position 1) = $\frac{2}{6}$ × $\frac{4}{5}$ × $\frac{3}{4}$ = $2\frac{4}{1}20$ = $\frac{1}{5}$ .
5.P(milk in position 2) = $\frac{4}{6}$ × $\frac{2}{5}$ × $\frac{3}{4}$ = $2\frac{4}{1}20$ = $\frac{1}{5}$ .
6.P(milk in position 3) = $\frac{4}{6}$ × $\frac{3}{5}$ × $\frac{2}{4}$ = $2\frac{4}{1}20$ = $\frac{1}{5}$ .
7.P(2 dark, 1 milk) = 3 × $\frac{1}{5}$ = $\frac{3}{5}$ .
8.P(more dark than milk) = $\frac{1}{5}$ + $\frac{3}{5}$ = $\frac{4}{5}$ .

Final answer

$\frac{4}{5}$

Mark points

M1: identifies the two qualifying cases (3 dark, or 2 dark and 1 milk).
M1: correctly computes P(3 dark) = $\frac{1}{5}$ using without-replacement fractions.
M1: sets up at least one correct route for P(2 dark, 1 milk) using without-replacement fractions.
M1: recognises and accounts for all three arrangements of 2 dark and 1 milk.
A1: $\frac{4}{5}$ .

Watch out

Forgetting that '2 dark, 1 milk' can occur in three different orders and counting only one arrangement, giving P = $\frac{1}{5}$ + $\frac{1}{5}$ = $\frac{2}{5}$ .

Grade 9 stretchproblem solving

A biased coin has P(Heads) = p. The coin is flipped twice. Given that P(exactly one head) = 0.42, find both possible values of p and explain in context which value is more plausible if the coin is described as 'slightly biased towards tails'.

6 marks6 minstree-g9

Show solution

Worked solution

1.P(exactly one head) = P(HT) + P(TH) = p(1−p) + (1−p)p = 2p(1−p).
2.Set 2p(1−p) = 0.42 → 2p − 2p² = 0.42 → 2p² − 2p + 0.42 = 0.
3.Divide by 2: p² − p + 0.21 = 0.
4.Quadratic formula: p = (1 ± $\sqrt{1 − 0.84}$ ) / 2 = (1 ± √0.16) / 2 = (1 ± 0.4) / 2.
5.p = 0.7 or p = 0.3.
6.Both are valid probabilities. 'Slightly biased towards tails' means P(Heads) < 0.5, so p = 0.3 is more plausible.

Final answer

p = 0.3 or p = 0.7. For a coin biased towards tails, p = 0.3 is more plausible.

Mark points

M1: correctly forms P(exactly one head) = 2p(1−p).
M1: sets up equation 2p(1−p) = 0.42 and rearranges to a quadratic.
M1: correctly applies quadratic formula or factorises to find both roots.
A1: both values p = 0.3 and p = 0.7 stated.
A1: selects p = 0.3 with a valid contextual reason (tails bias means p < 0.5).
A1 (communication): full clear explanation including the symmetry argument — p and 1−p give the same product, so two solutions are expected.

Watch out

Giving only one of the two valid values, or selecting the wrong value when explaining the contextual interpretation.

Exam Techniqueexam trap

A bag has 3 red and 2 blue balls. One ball is drawn, its colour noted, and it is NOT replaced. A second ball is then drawn. A student draws the following tree diagram and writes P(red second) = $\frac{3}{5}$ on both second-level red branches. Explain the error and give the two correct probabilities for red on the second draw.

3 marks3 minstree-et1

Show solution

Worked solution

1.The error: after drawing a red ball first, there are only 2 red balls and 4 balls in total remaining.
2.The probability of red on the second draw depends on what happened first — so it cannot be the same on both branches.
3.If red was drawn first: P(red second) = $\frac{2}{4}$ = $\frac{1}{2}$ .
4.If blue was drawn first: P(red second) = $\frac{3}{4}$ .
5.These are conditional probabilities — the denominator changes because sampling is without replacement.

Final answer

If red first: P(red second) = $\frac{2}{4}$ . If blue first: P(red second) = $\frac{3}{4}$ . The student incorrectly used the original probability $\frac{3}{5}$ on both branches.

Mark points

B1: identifies that the second-draw probabilities must change because sampling is without replacement.
M1: states $\frac{2}{4}$ (or equivalent) as the correct probability of red on the second draw given red was drawn first.
A1: states $\frac{3}{4}$ (or equivalent) as the correct probability of red on the second draw given blue was drawn first.

Watch out

Using the original bag's probabilities ( $\frac{3}{5}$ and $\frac{2}{5}$ ) on every second-level branch rather than updating the counts after the first draw.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Tree diagrams - 2 marksSame bag. Find P(both blue).Mark answer

Answer

$\frac{6}{1}0$ × $\frac{5}{9}$ = $3\frac{0}{9}0$ = $\frac{1}{3}$

2Probability basics - 2 marksP(A) = 0.35. Find P(not A).Mark answer

Answer

0.65

3Relative frequency - 2 marksExpected frequency of an outcome with probability 0.4 in 250 trials.Mark answer

Answer

100

4Sample spaces and frequency trees - 3 marksP(different colours) when drawing twice with replacement from the bag above.Mark answer

Answer

2 × ( $\frac{4}{6}$ × $\frac{2}{6}$ ) = $\frac{8}{1}8$ = $\frac{4}{9}$

Mastery check

I can recall the multiply-along / add-routes rule and explain why each operation is used at each stage.
I can draw a complete tree diagram for two-stage experiments, ensuring each pair of branches sums to 1.
I can correctly update second-stage probabilities in without-replacement problems by reducing both the total and the specific-colour count.
I can apply the complement rule (1 − P(none)) to efficiently solve 'at least one' problems without listing every route.
I can model an unfamiliar probability scenario — such as one involving a quadratic equation for an unknown probability — using the multiply-and-add structure of tree diagrams.

Ask Nova Bot

Draw and explain a tree diagram for rolling a die and then flipping a coin, labelling every branch with its probability and checking all branches sum to 1.I don't understand why probabilities change in a without-replacement problem — can you give me a hint using a small example?Quiz me with five tree-diagram questions: start easy (with replacement) and build up to a hard without-replacement 'at least one' question. Mark my answers.What are the three most common mistakes students make on tree-diagram GCSE exam questions, and how do I avoid each one?Show me a real-world situation where tree diagrams are used — for example in medical testing or weather forecasting — and explain the calculation.

Get help with anything that still feels tricky.

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