1MA1 · P6, P8ProbabilityFoundation & HigherGrades 4–7

Tree Diagrams & Combined Events

Q: How do I find P(at least one)?

Use 1 - P(none), which is usually quicker than adding every path with at least one.

What is the chance of rolling a six and then flipping a head? Of picking two red counters in a row? When events happen in sequence, a $\textbf{tree diagram}$ organises every possibility, and two rules do the rest: multiply along the branches for "and", and add the end results for "or".

50 min Probability

The big picture

For independent events — where one outcome does not affect the other — P(A and B) $=$ P(A) × P(B). Tree diagrams apply this repeatedly: each branch shows a probability, each path through the tree is a combined outcome, and multiplying along a path gives its probability.

The key Edexcel distinction is replacement. If a counter is put back, the second pick has the same probabilities as the first. If it is not, the numbers on the second set of branches change — both the total and the count of the colour removed.

What you'll learn

Use the addition rule for mutually exclusive events
Use the multiplication rule for independent events
Draw and use tree diagrams for two or three events
Solve problems with and without replacement

The OR rule

For mutually exclusive events (which cannot both happen), P(A or B) $=$ P(A) $+$ P(B).

Rolling a dice: P(2 or 5) $= \tfrac16 + \tfrac16 = \tfrac13$ .

P(A \text{ or } B) = P(A) + P(B)

Mutually exclusive events.

Your turnSame method, new numbers — try it before moving on.

Events $A$ and $B$ cannot both happen. $P(A) = 0.35$ and $P(B) = 0.2$ . Find $P(A \text{ or } B)$ .

Your answer

Show worked solution

Solution

1Mutually exclusive: add.
2 $0.35 + 0.2 = 0.55$ .

Answer: $0.55$

The AND rule

Events are $\textbf{independent}$ if the outcome of one does not affect the other.

For independent events, P(A and B) $=$ P(A) × P(B). Rolling a six and flipping a head: $\tfrac16 \times \tfrac12 = \tfrac{1}{12}$ .

Multiplying makes probabilities smaller, which makes sense: both things happening is less likely than either one.

P(A \text{ and } B) = P(A) \times P(B)

Independent events.

Your turnSame method, new numbers — try it before moving on.

The probability of passing test 1 is 0.7 and test 2 is 0.8, independently. Find the probability of passing both.

Your answer

Show worked solution

Solution

1Independent: multiply.
2 $0.7 \times 0.8 = 0.56$ .

Answer: $0.56$

Tip — Think "AND means multiply, OR means add". But only add outcomes that cannot happen together.

Tree diagrams

Draw one set of branches for each event. Each set of branches from a point must add to 1.

Multiply along each path to get the probability of that combined outcome. The end probabilities add to 1.

To find a probability like "one of each", add the probabilities of all paths that fit.

Example 1A bag has 5 red and 3 blue counters. A counter is taken, its colour noted and it is replaced, then a second counter is taken. Find P(both the same colour).

1Branches: red

\tfrac58

, blue

\tfrac38

, for both picks.

2P(red, red)

= \tfrac{25}{64}

; P(blue, blue)

= \tfrac{9}{64}

3Add:

\tfrac{34}{64}

Answer

\tfrac{17}{32}

Your turnSame method, new numbers — try it before moving on.

A bag has 4 red and 6 blue counters. A counter is taken and replaced, then a second is taken. Find the probability of one of each colour.

Your answer

Show worked solution

Solution

1RB: $0.4 \times 0.6 = 0.24$ ; BR: $0.6 \times 0.4 = 0.24$ .
2Add: $0.48$ .

Answer: $0.48$

Without replacement

If the first item is not replaced, there is one fewer item for the second pick, and one fewer of the colour taken.

The second-pick probabilities depend on the first result, so the events are $\textbf{dependent}$ .

Example 1The same bag of 5 red and 3 blue counters; two counters are taken without replacement. Find P(both the same colour).

1P(red, red)

= \tfrac58 \times \tfrac47 = \tfrac{20}{56}

2P(blue, blue)

= \tfrac38 \times \tfrac27 = \tfrac{6}{56}

3Add:

\tfrac{26}{56}

Answer

\tfrac{13}{28}

Your turnSame method, new numbers — try it before moving on.

The same bag of 4 red and 6 blue counters; two are taken without replacement. Find the probability both are blue.

Your answer

Show worked solution

Solution

1 $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$ .
2 $= \frac{1}{3}$ .

Answer: $\frac{1}{3}$

Why this works

Without replacement, getting two of the same colour is slightly less likely ( $\tfrac{13}{28} \approx 0.464$ ) than with replacement ( $\tfrac{17}{32} \approx 0.531$ ), because removing a red makes a second red harder to get.

Think like an examiner

Label every branch with its probability.
Show multiplication along paths and addition of paths.
Adjust second-stage probabilities without replacement.

Combined events

P(A \text{ or } B) = P(A) + P(B)

Mutually exclusive.

P(A \text{ and } B) = P(A) \times P(B)

Independent.

\text{branches from a point sum to } 1

Tree diagram check.

Watch out for these

Adding along branches.

Multiply along a path; add different paths.

Keeping the denominator 8 on second picks without replacement.

Use 7 for the second pick.

Forgetting one of the two paths for "one of each".

Red then blue and blue then red both count.

Multiplying probabilities of events that are not independent without adjusting.

Use the conditional probabilities on the second branches.

Stretch yourself

A bag contains $n$ sweets, of which 6 are orange. Two sweets are taken at random without replacement. The probability that both are orange is $\tfrac13$ . Show that $n^{2} - n - 90 = 0$ and find $n$ .

Hint — Write P(orange, orange) as a product of two fractions in terms of $n$ .

Reveal solution

SolutionP(both orange)

= \dfrac{6}{n} \times \dfrac{5}{n - 1} = \dfrac{30}{n(n - 1)}

. Setting this equal to

\tfrac13

n(n - 1) = 90

, so

n^{2} - n - 90 = 0

. Factorising,

(n - 10)(n + 9) = 0

, and since

n

is positive,

n = 10

. Check:

\tfrac{6}{10} \times \tfrac59 = \tfrac{30}{90} = \tfrac13

✓.

\blacksquare

Questions students ask

How do I know whether events are independent?

If one outcome changes the probabilities of the other — like not replacing a counter — they are dependent. Coin flips and dice rolls are independent.

Do I always need a tree diagram?

No, but it helps avoid missing paths, especially for "at least one" or "one of each".

How do I find P(at least one)?

Use

1 -

P(none), which is usually quicker than adding every path with at least one.

Key takeaways

OR (mutually exclusive): add. AND (independent): multiply.
Tree branches from each point sum to 1.
Multiply along paths; add the paths that fit the event.
Without replacement, second-stage probabilities change.
P(at least one) $= 1 -$ P(none).

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

The probability of rain on Saturday is 0.3 and on Sunday is 0.4, independently. Work out the probability that it rains on neither day.[2]
Show mark scheme
- M1 $0.7 \times 0.6$
- A1 $0.42$
Answer: $0.42$

Question 2Grade 63 marks

A bag has 7 green and 3 yellow sweets. Two are taken at random without replacement. Work out the probability that at least one is yellow.[3]
Show mark scheme
- M1 $P(GG) = \frac{7}{10} \times \frac{6}{9}$
- M1 $1 - \frac{42}{90}$
- A1 $\frac{8}{15}$
Answer: $\frac{8}{15}$

Question 3Grade 73 marks

A biased coin is flipped twice. The probability of two heads is 0.36.

Work out the probability of exactly one head.[3]
Show mark scheme
- M1 $P(H) = \sqrt{0.36} = 0.6$
- M1 $2 \times 0.6 \times 0.4$
- A1 $0.48$
Answer: $0.48$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

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