Tree Diagrams · Study Notes · Edexcel IGCSE 4MA1 Higher

Tree diagrams visualise sequential probability. Multiply along branches, add the relevant paths. With/without replacement is a guaranteed Higher Tier topic.

At a glance

Tree = branching diagram showing all outcomes.
Each branch labelled with a probability.
Multiply along a path (AND).
Add the path probabilities for compound events (OR).
With replacement: probabilities unchanged → independent.
Without replacement: probabilities CHANGE → conditional.
Branches from any node must sum to 1.
'At least one' → use complement.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

6.5 — Construct a tree diagram for a two-stage event.
6.5 — Calculate probabilities by multiplying along branches and adding outcomes.
6.5 — Solve problems with replacement and without replacement.

Building a tree diagram

▼

Each branch = an outcome. Each branch labelled with its probability.

A tree diagram shows all outcomes of a sequence of events.

Structure:

Start from a single point.
First "level" of branches: outcomes of the first event.
From each first-level node, draw second-level branches: outcomes of the second event.
Each branch has a probability label.

Example. Two coins.

       0.5  H ─── 0.5  H  (HH)
            │     0.5  T  (HT)
            │
       0.5  T ─── 0.5  H  (TH)
                  0.5  T  (TT)

Rule. Probabilities of branches from any single node SUM to 1.

Worked example. Bag: 3 red, 5 blue. Pick two WITHOUT replacement.

First level: $P(R_1) = 3/8$ , $P(B_1) = 5/8$ .

After picking red: 2 red and 5 blue (7 total) remain. After picking blue: 3 red and 4 blue (7 total) remain.

Second level:

After $R_1$ : $P(R_2|R_1) = 2/7$ , $P(B_2|R_1) = 5/7$ .
After $B_1$ : $P(R_2|B_1) = 3/7$ , $P(B_2|B_1) = 4/7$ .

Note: the second-level probabilities DEPEND on what happened first. With replacement, they would all be the same.

Without replacement: the second pick has one fewer counter, so its probabilities change.

Edexcel tip. Lay out the tree clearly, with probabilities written ON the branches. Mark schemes credit a fully-labelled tree even before computation.

Branches show outcomes.
Each branch = probability.
Branches from a node sum to 1.
Without replacement → second probabilities change.

Multiply along branches; add the paths

▼

AND multiplies down. OR adds across.

Two rules:

Multiply along a path (AND): the probability of a single full outcome is the product of probabilities along its branches.
Add the path probabilities (OR): if multiple paths give the desired outcome, sum their probabilities.

Multiply probabilities along each path, then add the paths that give the wanted outcome.

Worked example. Bag: 3 red, 5 blue. Without replacement. $P(\text{exactly one red})$ ?

Two paths give exactly one red: $R_1B_2$ and $B_1R_2$ .

$P(R_1B_2) = (3/8)(5/7) = 15/56$ .
$P(B_1R_2) = (5/8)(3/7) = 15/56$ .
Total: $15/56 + 15/56 = 30/56 = 15/28$ .

Worked example. Same bag. $P(\text{both same colour})$ ?

Two paths: $R_1R_2$ and $B_1B_2$ .

$P(R_1R_2) = (3/8)(2/7) = 6/56 = 3/28$ .
$P(B_1B_2) = (5/8)(4/7) = 20/56 = 5/14$ .
Total: $3/28 + 10/28 = 13/28$ .

Worked example. $P(\text{at least one red})$ ?

Direct: many paths. EASIER via complement: $P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - 5/14 = 9/14$ .

Edexcel tip. "At least one" almost always uses complement. Save time.

Multiply along branches.
Add path probabilities.
'Exactly one' → typically two paths.
'At least one' → use complement.

See the full worked example for tree diagrams →

With vs without replacement

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With: independent. Without: probabilities change.

With replacement. After the first pick, the item is put BACK. Total stays the same. Probabilities for the second pick are the SAME as the first.

Worked example. Bag: 4 red, 6 blue. With replacement. $P(\text{both red})$ ?

$P(R_1) = 4/10$ . After replacing, $P(R_2) = 4/10$ .
$P(\text{both red}) = (4/10)(4/10) = 16/100 = 4/25$ .

Without replacement. After the first pick, the item is NOT put back. Total decreases. Whatever was picked is unavailable.

Worked example. Same bag, without replacement.

$P(R_1) = 4/10$ . After taking a red: 3 red, 6 blue, 9 total. $P(R_2|R_1) = 3/9 = 1/3$ .
$P(\text{both red}) = (4/10)(3/9) = 12/90 = 2/15$ .

Spot the difference. With replacement gives $4/25 \approx 0.16$ . Without gives $2/15 \approx 0.133$ . Without is smaller because the second probability is reduced.

Edexcel tip. Read the question carefully. Keywords:

"With replacement" / "replaces" → independent, probabilities unchanged.
"Without replacement" / "does not replace" → conditional, probabilities reduced.

If the situation is something like "two students are chosen at random" — without replacement (you can't pick the same student twice).

With: probabilities unchanged.
Without: probabilities reduce.
Read the question carefully.
'Two students chosen' → usually without replacement.

Quick recap

Tree shows outcomes; each branch labelled with probability.
Multiply along a path; add path probabilities for compound events.
Branches from a node sum to 1.
With replacement: independent. Without: conditional.
'At least one' → complement.
'Exactly one' → typically two paths to add.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Multiply along branches (AND).
Add the path probabilities (OR).
Without replacement: reduce both numerator and denominator.
Branches at a node sum to 1.
$P(\text{at least one}) = 1 - P(\text{none})$ .

How it’s examined

Tree diagrams + without replacement is THE classic Higher Tier 6-mark question. Examiner reports flag: (1) wrong second-pick denominator, (2) only computing one path when two are needed, (3) not using complement for 'at least one'.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on tree diagrams, written exactly the way a tutor would explain them at the board.

1Tree with replacement (independent)
Higher• tree, with replacement
▼
Question
Bag: 4 red, 6 blue. Pick two WITH replacement. Find $P(\text{both red})$ .
Step-by-step solution
1. Step 1
  $P(R) = 4/10 = 0.4$ each pick (independent).
2. Step 2
  Multiply along R-R branch: $0.4 \times 0.4 = 0.16$ .
Answer
$P(\text{both red}) = 0.16$
2Tree without replacement
Higher• Adapted from 4MA1/2H May/Jun 2024 Q19• tree, without replacement
▼
Question
Bag: 3 red, 5 blue. Pick two WITHOUT replacement. Find $P(\text{exactly one red})$ .
Step-by-step solution
1. Step 1
  1st red probabilities: $3/8$ . 1st blue: $5/8$ .
2. Step 2
  After red taken: 2 red, 5 blue. After blue: 3 red, 4 blue.
3. Step 3
  Two paths give exactly one red: RB and BR.
4. Step 4
  $P(\text{RB}) = (3/8)(5/7) = 15/56$ . $P(\text{BR}) = (5/8)(3/7) = 15/56$ .
5. Step 5
  Total: $15/56 + 15/56 = 30/56 = 15/28$ .
Answer
$P(\text{exactly one red}) = 15/28$
Examiner tip
Multiply along branches; add the relevant outcomes. Mark scheme: M1 for tree with correct probabilities, M1 for multiplying, M1 for adding paths, A1 for value.
3At least one (use complement)
Higher• complement, tree
▼
Question
Bag: 3 red, 5 blue. Pick two without replacement. Find $P(\text{at least one red})$ .
Step-by-step solution
1. Step 1
  Complement: $P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - P(\text{both blue})$ .
2. Step 2
  $P(\text{both blue}) = (5/8)(4/7) = 20/56 = 5/14$ .
3. Step 3
  $1 - 5/14 = 9/14$ .
Answer
$P(\text{at least one red}) = 9/14$
Examiner tip
'At least one' problems → use complement to save time.
4Spotting an impossible pair of branches [1 mark]
Getting started• Paper 1H/2H style — 1 mark• tree diagrams, probability
▼
Question
Tom draws a tree diagram for whether a netball player scores or misses a shot. On the pair of branches for the first shot he writes Scores $0.4$ and Misses $0.5$ .

Explain why Tom's probabilities must be wrong. [1]
Step-by-step solution
1. Step 1
  Scores and Misses are the only two outcomes of a shot, and exactly one of them happens.
2. Step 2
  B1 So the two probabilities must add up to $1$ , but Tom's add up to $0.9$ .
  $0.4 + 0.5 = 0.9 \neq 1$
Answer
The branches from one point must add up to $1$ , and $0.4 + 0.5 = 0.9$
Examiner tip
Examiner tip: An explanation needs a mathematical reason with the numbers. Say what the total should be and what Tom's total actually is.

Common mistake: Writing "the numbers are too small" or "he has done it wrong". Neither mentions the total of $1$ , so neither scores.
5Three letter tiles in a given order, with replacement [2 marks]
Getting started• Paper 1H/2H style — 2 marks• tree diagrams, with replacement
▼
Question
Six letter tiles spell the word BANANA. The tiles are put in a bag. Nadia takes a tile at random, writes down its letter and puts the tile back. She does this three times.

Work out the probability that she writes down N, then A, then N, in that order. [2]
Step-by-step solution
1. Step 1
  Count the tiles, not the letters: $2$ of the $6$ tiles show N and $3$ show A. Each tile is put back, so the denominator stays $6$ on every pick.
  $P(\text{N}) = \frac{2}{6}, \quad P(\text{A}) = \frac{3}{6}$
2. Step 2
  M1 The order is fixed, so there is only one path: multiply along N–A–N.
  $\frac{2}{6} \times \frac{3}{6} \times \frac{2}{6} = \frac{12}{216}$
3. Step 3
  A1 Simplify.
  $= \frac{1}{18}$
Answer
$\frac{1}{18}$
Examiner tip
Examiner tip: "Puts the tile back" means with replacement: every stage of the tree has the same branches. "In that order" means one path only, so do not multiply by the number of arrangements.

Common mistake: Using $\frac{1}{3}$ for each letter because there are three different letters (B, A, N); the tiles are not equally split between the letters. Using $\frac{2}{6} \times \frac{3}{5} \times \frac{1}{4}$ is the without-replacement calculation.
6Neither day, when the second day depends on the first [2 marks]
Getting started• Paper 1H/2H style — 2 marks• tree diagrams, conditional probability
▼
Question
The probability that it snows in a mountain village on a Monday in January is $0.2$ .

If it snows on Monday, the probability that it snows on Tuesday is $0.6$ . If it does not snow on Monday, the probability that it snows on Tuesday is $0.1$ .

Work out the probability that it does not snow on either day. [2]
Step-by-step solution
1. Step 1
  No snow on Monday has probability $1 - 0.2 = 0.8$ . The Tuesday branch then comes from the no-snow half of the tree, where no snow has probability $1 - 0.1 = 0.9$ .
2. Step 2
  M1 Multiply along the No snow–No snow path.
  $0.8 \times 0.9$
3. Step 3
  A1 Evaluate.
  $= 0.72$
Answer
$0.72$
Examiner tip
Examiner tip: When the second-day probabilities depend on the first day, write both pairs of second branches before you choose one. Each pair adds up to $1$ .

Common mistake: Using $1 - 0.6 = 0.4$ for Tuesday, which is the no-snow branch after a snowy Monday, giving $0.8 \times 0.4 = 0.32$ . Another error is $1 - 0.2 \times 0.6 = 0.88$ , which is the probability that it does not snow on both days, a different event.
7Probability of a faulty item from two machines [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• tree diagrams, conditional probability
▼
Question
A factory has two machines, A and B. Machine A makes $60\%$ of the items and machine B makes the rest.

The probability that an item made by machine A is faulty is $0.03$ . The probability that an item made by machine B is faulty is $0.05$ .

An item is chosen at random. Work out the probability that it is faulty. [3]
Step-by-step solution
1. Step 1
  First-stage branches: A $0.6$ , B $0.4$ . Second-stage branches: after A, Faulty $0.03$ ; after B, Faulty $0.05$ .
2. Step 2
  M1 Multiply along one path to Faulty.
  $0.6 \times 0.03 = 0.018$
3. Step 3
  M1 Multiply along the other path to Faulty and add.
  $0.4 \times 0.05 = 0.020, \qquad 0.018 + 0.020$
4. Step 4
  A1 Complete the addition.
  $= 0.038$
Answer
$0.038$
Examiner tip
Examiner tip: Change the percentage to a probability ( $60\% = 0.6$ ) before it goes on a branch, and remember that machine B's branch is $1 - 0.6 = 0.4$ .

Common mistake: Adding $0.03 + 0.05 = 0.08$ or averaging to $0.04$ . That ignores the fact that machine A makes more of the items.
8Two different colours from a bag of three colours [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• tree diagrams, without replacement, complement
▼
Question
A bag contains $2$ red counters, $3$ green counters and $4$ yellow counters. Ella takes two counters at random from the bag without replacement.

Work out the probability that the two counters are different colours. [3]
Step-by-step solution
1. Step 1
  There are six paths with different colours but only three with the same colour, so find P(same colour) and subtract from $1$ .
2. Step 2
  M1 Work out the three same-colour paths.
  $\frac{2}{9} \times \frac{1}{8} + \frac{3}{9} \times \frac{2}{8} + \frac{4}{9} \times \frac{3}{8} = \frac{2 + 6 + 12}{72} = \frac{20}{72}$
3. Step 3
  M1 Subtract from $1$ .
  $1 - \frac{20}{72} = \frac{52}{72}$
4. Step 4
  A1 Simplify.
  $= \frac{13}{18}$
Answer
$\frac{13}{18}$
Examiner tip
Examiner tip: With three colours, a tree has nine end points. Count the paths for each event and work out the shorter list, then use the complement if needed.

Common mistake: Adding only three of the six different-colour paths (for example RG, RY and GY), which counts each pair in one order only and gives half the probability of those pairs.
9Exactly two of three people hit a target [4 marks]
Stretch• Paper 1H/2H style — 4 marks• tree diagrams, independent events
▼
Question
Three archers each shoot one arrow at a target. The probabilities that they hit the target are $0.9$ , $0.8$ and $0.7$ . The results are independent.

Work out the probability that exactly two of the three archers hit the target. [4]
Step-by-step solution
1. Step 1
  M1 The one archer who misses can be the first, the second or the third. Work out one of these paths, using $1 - p$ for the miss.
  $P(\text{H, H, M}) = 0.9 \times 0.8 \times 0.3 = 0.216$
2. Step 2
  M1 Work out the other two paths. The probabilities are all different, so each path must be calculated.
  $P(\text{H, M, H}) = 0.9 \times 0.2 \times 0.7 = 0.126, \quad P(\text{M, H, H}) = 0.1 \times 0.8 \times 0.7 = 0.056$
3. Step 3
  M1 Add the three paths.
  $0.216 + 0.126 + 0.056$
4. Step 4
  A1 Evaluate.
  $= 0.398$
Answer
$0.398$
Examiner tip
Examiner tip: Only use " $3 \times$ one path" when all the paths have the same probability. Here each archer has a different probability, so all three products are needed.

Common mistake: Writing $3 \times 0.216 = 0.648$ . The three arrangements are all needed, but they are not equally likely here.
10Finding the number of blue counters from an "at least one" probability [5 marks]
Stretch• Paper 1H/2H style — 5 marks• tree diagrams, without replacement, complement, quadratic
▼
Question
A bag contains $4$ red counters and $n$ blue counters. Two counters are taken at random from the bag without replacement.

The probability that at least one of the counters is red is $\frac{14}{15}$ .

Work out the value of $n$ . Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 The only path with no red counter is blue then blue. There are $n + 4$ counters, then $n + 3$ after one blue counter is taken.
  $P(\text{B, B}) = \frac{n}{n + 4} \times \frac{n - 1}{n + 3}$
2. Step 2
  M1 At least one red is everything except blue–blue, so $P(\text{B, B}) = 1 - \frac{14}{15} = \frac{1}{15}$ . Set up the equation and clear the fractions.
  $\frac{n(n - 1)}{(n + 4)(n + 3)} = \frac{1}{15} \;\Rightarrow\; 15n(n - 1) = (n + 4)(n + 3)$
3. Step 3
  A1 Expand and rearrange to a quadratic equal to zero.
  $15n^2 - 15n = n^2 + 7n + 12 \;\Rightarrow\; 14n^2 - 22n - 12 = 0 \;\Rightarrow\; 7n^2 - 11n - 6 = 0$
4. Step 4
  M1 Factorise (or use the quadratic formula).
  $(7n + 3)(n - 2) = 0 \;\Rightarrow\; n = -\frac{3}{7} \text{ or } n = 2$
5. Step 5
  A1 The number of counters must be a positive whole number.
  $n = 2$
Answer
$n = 2$
Examiner tip
Examiner tip: Turn "at least one red" into its complement first: the equation is much simpler for the single path blue–blue than for the three paths that contain a red. Check in context: with $4$ red and $2$ blue, $1 - \frac{2}{6} \times \frac{1}{5} = 1 - \frac{1}{15} = \frac{14}{15}$ .

Common mistake: Setting the blue–blue expression equal to $\frac{14}{15}$ instead of $\frac{1}{15}$ . Trying values of $n$ until one works does not earn the method marks when the question says "Show clear algebraic working".

Question 1
Paper 1H/2H style — 1 mark1 mark
Two counters are taken at random from a bag, one after the other, without replacement.

Explain why the events "the first counter is red" and "the second counter is red" are not independent. [1]
Model answer
The first counter is not put back, so the colour of the first counter changes the number of red counters left for the second pick. The probability that the second counter is red depends on what happened first.
Why this scores
Mark scheme:
- B1 for a statement that the probability for the second counter depends on (changes with) the first counter oe
Why this scores full marks: The answer links "not independent" to a change in probability, and gives the reason (no replacement). Saying only "because there is no replacement" does not explain what that does to the probability.
Question 2
Paper 1H/2H style — 2 marks2 marks
The probability that a seed grows into a plant is $0.85$ . If the seed grows into a plant, the probability that the plant flowers is $0.6$ . A seed that does not grow cannot flower.

Suki plants one seed. Work out the probability that it does not produce a flower. [2]
Model answer
The tree stops after Does not grow. A flower needs Grows then Flowers:

$0.85 \times 0.6 = 0.51$

$1 - 0.51 = 0.49$

Answer: $0.49$
Why this scores
Mark scheme:
- M1 for $1 - 0.85 \times 0.6$ or $0.15 + 0.85 \times 0.4$ oe
- A1 for $0.49$ oe
Why this scores full marks: The candidate sees that "no flower" happens on two paths (does not grow, or grows but does not flower) and uses the one path that gives a flower instead. Giving $0.4$ or $0.85 \times 0.4 = 0.34$ misses the seeds that never grow.
Question 3
Paper 1H/2H style — 2 marks2 marks
A pot contains $7$ red pens and $3$ blue pens. Chen takes a pen at random and does not replace it. He then takes a second pen at random.

Work out the probability that the first pen is red and the second pen is blue. [2]
Model answer
$\frac{7}{10} \times \frac{3}{9} = \frac{21}{90}$

Answer: $\frac{7}{30}$
Why this scores
Mark scheme:
- M1 for $\frac{7}{10} \times \frac{3}{9}$
- A1 for $\frac{7}{30}$ oe
Why this scores full marks: The question fixes the order (red first, then blue), so only one path is needed. Doubling it would give the probability of one of each colour in either order, which is not what is asked.
Question 4
Paper 1H/2H style — 3 marks3 marks
Aiden takes two penalties. The probability that he scores is the same for each penalty, and the penalties are independent.

The probability that he scores both penalties is $0.49$ .

Work out the probability that he scores exactly one of the two penalties. [3]
Model answer
Let $p$ be the probability that he scores one penalty.

$p^2 = 0.49 \;\Rightarrow\; p = 0.7$

$0.7 \times 0.3 + 0.3 \times 0.7 = 0.21 + 0.21$

Answer: $0.42$
Why this scores
Mark scheme:
- M1 for $p = \sqrt{0.49}$ or $p = 0.7$
- M1 for $2 \times p \times (1 - p)$ with their $p$ oe
- A1 for $0.42$ oe
Why this scores full marks: The candidate works backwards from "both" to a single branch, then forwards again. Halving $0.49$ to get $p$ is a common error: "both" means multiply, so the reverse is the square root.
Question 5
Paper 1H/2H style — 3 marks3 marks
A bag contains $5$ red counters and $5$ green counters. Two counters are taken at random from the bag.

Show that the probability that both counters are the same colour is smaller when the first counter is not replaced than when it is replaced. [3]
Model answer
With replacement:

$\frac{5}{10} \times \frac{5}{10} + \frac{5}{10} \times \frac{5}{10} = \frac{50}{100} = \frac{1}{2}$

Without replacement:

$\frac{5}{10} \times \frac{4}{9} + \frac{5}{10} \times \frac{4}{9} = \frac{40}{90} = \frac{4}{9}$

$\frac{4}{9} = \frac{8}{18}$ and $\frac{1}{2} = \frac{9}{18}$ , so $\frac{4}{9} < \frac{1}{2}$ .

The probability is smaller without replacement.
Why this scores
Mark scheme:
- M1 for $\frac{5}{10} \times \frac{5}{10} + \frac{5}{10} \times \frac{5}{10}$ oe or $\frac{1}{2}$
- M1 for $\frac{5}{10} \times \frac{4}{9} + \frac{5}{10} \times \frac{4}{9}$ oe
- A1 for $\frac{1}{2}$ and $\frac{4}{9}$ with a correct comparison and conclusion
Why this scores full marks: Both probabilities are worked out in full and then compared with a common denominator, and the final sentence answers the question. A "show that" needs the conclusion written, not left for the examiner to infer.
Question 6
Paper 1H/2H style — 4 marks4 marks
A fair spinner has three equal sections: one red and two white. Ola spins the spinner three times.

(a) Work out the probability that the spinner does not land on red in any of the three spins. [1]

(b) Work out the probability that the spinner lands on red exactly twice. [3]
Model answer
(a)

$\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$

Answer: $\frac{8}{27}$

(b) RRW, RWR, WRR

$\frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}$

$3 \times \frac{2}{27} = \frac{6}{27}$

Answer: $\frac{2}{9}$
Why this scores
Mark scheme:
- B1 for $\frac{8}{27}$ oe
- M1 for $\frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}$
- M1 for $3 \times \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}$ oe
- A1 for $\frac{2}{9}$ oe
Why this scores full marks: The three arrangements are listed before the multiplying. Stopping at $\frac{2}{27}$ uses one arrangement only; examiners report that many candidates find one arrangement but do not realise there are three.
Question 7
Paper 1H/2H style — 4 marks4 marks
Priya throws darts at a target. The probability that she hits the bullseye with any dart is $0.3$ , independently of other throws. She stops as soon as she hits the bullseye, or after three throws, whichever comes first.

(a) Work out the probability that Priya throws exactly two darts. [2]

(b) Work out the probability that Priya hits the bullseye. [2]
Model answer
(a) Miss then hit:

$0.7 \times 0.3 = 0.21$

Answer: $0.21$

(b) She fails to hit only if she misses all three:

$1 - 0.7 \times 0.7 \times 0.7 = 1 - 0.343$

Answer: $0.657$
Why this scores
Mark scheme:
- M1 for $0.7 \times 0.3$
- A1 for $0.21$ oe
- M1 for $1 - 0.7^3$ or $0.3 + 0.7 \times 0.3 + 0.7^2 \times 0.3$
- A1 for $0.657$ oe
Why this scores full marks: The candidate reads the stopping rule: throwing exactly two darts means the first missed and the second hit. In (b) the complement avoids adding three separate paths; the direct sum $0.3 + 0.21 + 0.147$ gives the same $0.657$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
A bag contains $3$ red counters and $5$ blue counters. Leah takes a counter at random from the bag. She puts the counter back in the bag together with another counter of the same colour. She then takes a second counter at random from the bag.

Work out the probability that the two counters she takes are the same colour. [4]
Model answer
After the first counter there are $9$ counters in the bag. After red: $4$ red, $5$ blue. After blue: $3$ red, $6$ blue.

$P(\text{R, R}) = \frac{3}{8} \times \frac{4}{9} = \frac{12}{72}$

$P(\text{B, B}) = \frac{5}{8} \times \frac{6}{9} = \frac{30}{72}$

$\frac{12}{72} + \frac{30}{72} = \frac{42}{72}$

Answer: $\frac{7}{12}$
Why this scores
Mark scheme:
- M1 for $\frac{4}{9}$ or $\frac{6}{9}$ as a second-stage probability
- M1 for $\frac{3}{8} \times \frac{4}{9}$ or $\frac{5}{8} \times \frac{6}{9}$
- M1 for $\frac{3}{8} \times \frac{4}{9} + \frac{5}{8} \times \frac{6}{9}$
- A1 for $\frac{7}{12}$ oe
Why this scores full marks: The candidate works out the new contents of the bag before the second pick: the total goes up to $9$ and the colour just taken goes up by one. Treating the counter as simply replaced ( $\frac{3}{8} \times \frac{3}{8} + \frac{5}{8} \times \frac{5}{8} = \frac{34}{64}$ ) or as not replaced ( $\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{4}{7} = \frac{26}{56}$ ) misreads the question.
Question 9
Paper 1H/2H style — 5 marks5 marks
A box contains $10$ discs. $x$ of the discs are red and the rest are white. Two discs are taken at random from the box without replacement.

The probability that both discs are red is $\frac{1}{15}$ .

Work out the value of $x$ . Show clear algebraic working. [5]
Model answer
$\frac{x}{10} \times \frac{x - 1}{9} = \frac{1}{15}$

$15x(x - 1) = 90$

$x^2 - x - 6 = 0$

$(x - 3)(x + 2) = 0$

$x = 3$ or $x = -2$ , and $x$ cannot be negative.

Answer: $x = 3$
Why this scores
Mark scheme:
- M1 for $\frac{x}{10} \times \frac{x - 1}{9}$
- M1 for equating to $\frac{1}{15}$ and removing the fractions, e.g. $15x(x - 1) = 90$ oe
- A1 for $x^2 - x - 6 = 0$ oe
- M1 for $(x - 3)(x + 2)$ or a correct use of the quadratic formula
- A1 for $x = 3$ cao
Why this scores full marks: The candidate keeps the second-stage probability as $\frac{x - 1}{9}$ , because a red disc has gone. The negative root is rejected with a reason, and the final answer is a single value.
Question 10
Paper 1H/2H style — 5 marks5 marks
In a game, a player rolls a fair six-sided dice. If the dice shows $6$ , the player wins. If not, the player rolls the dice again and wins if it shows $5$ or $6$ . Otherwise the player loses.

(a) Show that the probability that a player wins the game is $\frac{4}{9}$ [3]

(b) Hana plays the game twice. Work out the probability that she wins exactly one of the two games. [2]
Model answer
(a) Win on the first roll, or not a $6$ then a $5$ or $6$ :

$\frac{1}{6} + \frac{5}{6} \times \frac{2}{6} = \frac{6}{36} + \frac{10}{36} = \frac{16}{36} = \frac{4}{9}$

(b)

$\frac{4}{9} \times \frac{5}{9} + \frac{5}{9} \times \frac{4}{9} = \frac{20}{81} + \frac{20}{81}$

Answer: $\frac{40}{81}$
Why this scores
Mark scheme:
- M1 for $\frac{5}{6} \times \frac{2}{6}$
- M1 for $\frac{1}{6} + \frac{5}{6} \times \frac{2}{6}$
- A1 for $\frac{16}{36}$ or better leading to $\frac{4}{9}$ with no errors
- M1 for $\frac{4}{9} \times \frac{5}{9} + \frac{5}{9} \times \frac{4}{9}$ oe
- A1 for $\frac{40}{81}$ oe
Why this scores full marks: The tree for one game stops after a $6$ on the first roll, so the "win" branch there has no second roll. In (b) the candidate uses the answer to (a) as a single branch and includes both orders, win–lose and lose–win.

4MA1 Tree Diagrams

At a glance

What you’ll learn

Building a tree diagram

Multiply along branches; add the paths

With vs without replacement

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Tree with replacement (independent)

2Tree without replacement

3At least one (use complement)

4Spotting an impossible pair of branches [1 mark]

5Three letter tiles in a given order, with replacement [2 marks]

6Neither day, when the second day depends on the first [2 marks]

7Probability of a faulty item from two machines [3 marks]

8Two different colours from a bag of three colours [3 marks]

9Exactly two of three people hit a target [4 marks]

10Finding the number of blue counters from an "at least one" probability [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10