Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Probability
Practise Tree diagrams and combined events. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Combined events are two or more events together, such as rolling two dice or taking two counters from a bag. You list outcomes systematically, use sample space diagrams and draw tree diagrams, multiplying along branches for 'and' and adding for 'or'. Independent events and simple 'without replacement' questions are on both tiers; longer three-event and algebraic problems appear on Higher papers at grades 6 to 9.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
List all the outcomes systematicallyFix the first item and list every option for the second, then change the first, e.g. HH, HT, TH, TT for two coins.
3
Use a sample space diagramDraw a grid for two dice or spinners, then count the outcomes you want out of the total number of cells.
4
Complete a tree diagram for independent eventsEach pair of branches adds up to 1, and the second set of branches repeats the first when the events are independent.
5
Multiply for 'and', add for 'or'Multiply along each path, then add the probabilities of all the paths that give the result you want.
5
Use tree diagrams without replacementFor the second pick the total goes down by 1, and so does the count of the colour already taken.
Notes
Listing outcomes and sample spaces
List outcomes systematically so you do not miss any: keep the first item fixed while you go through every option for the second.
A sample space diagram is a grid showing every outcome of two events. Two fair six-sided dice give 6 × 6 = 36 equally likely outcomes.
Count the cells you want: a total of 9 comes from 3 + 6, 4 + 5, 5 + 4 and 6 + 3, so P(total of 9) = \(\frac{4}{36} = \frac{1}{9}\).
This only works when every outcome is equally likely (fair dice, fair spinners).
Tree diagrams
Each set of branches shows one event, with a probability on each branch.
The branches coming from one point always add up to 1.
And: multiply along the branches to get the probability of that path.
Or: add the probabilities of the different paths. All the end-of-path probabilities add up to 1.
Example: P(rain) = 0.3 each day, independently. P(rain on exactly one of two days) = 0.3 × 0.7 + 0.7 × 0.3 = 0.21 + 0.21 = 0.42.
At least one is quickest as 1 − P(none): P(rain on at least one day) = 1 − 0.7 × 0.7 = 1 − 0.49 = 0.51.
Independent and dependent events
Independent: one event does not affect the other, e.g. the counter is put back. Then P(A and B) = P(A) × P(B), and every second set of branches is the same.
Dependent: the first outcome changes the probabilities for the second, e.g. taking counters without replacement, or eating a sweet.
Without replacement, the total goes down by 1 for the second pick. The count for a colour goes down only if that colour was taken first.
Example: 5 red and 3 blue counters, two taken without replacement. P(both red) = \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\).
Using the same probabilities on every set of branches assumes the events are independent: you may be asked to state this.
Cheatsheet
And: multiply along the branches
Or: add the probabilities of the separate paths
The branches from one point add up to 1
All the end-of-path probabilities add up to 1
Independent events: P(A and B) = P(A) × P(B)
Without replacement: the total goes down by 1 for each pick
Two fair six-sided dice: 36 equally likely outcomes
At least one = 1 − P(none)
How to answer each type of question
Use a sample space diagram
2 marks3
Draw a grid: one event across the top, the other down the side.
Fill in every cell (the total, the product or the pair of outcomes).
Count the cells you want and write that number over the total number of cells.
Example. Spinner A is numbered 1, 2 and 3. Spinner B is numbered 1, 2, 3 and 4. Both spinners are fair. Kai spins both spinners and multiplies the two numbers. Work out the probability that the product is greater than 5.
Show the model answer
Products (M1 for all 12 outcomes shown): A = 1: products 1, 2, 3, 4 A = 2: products 2, 4, 6, 8 A = 3: products 3, 6, 9, 12 Greater than 5: 6, 8, 6, 9 and 12, which is 5 out of 12. P(product greater than 5) = \(\frac{5}{12}\) (A1)
Complete a tree diagram and find probabilities (independent events)
4 to 6 marks4
Write the missing branch probabilities: each pair adds up to 1.
For independent events, copy the first set of probabilities onto every second set.
Multiply along a path for 'and'.
Add paths for 'or': 'exactly one' is late then on time, plus on time then late.
Example. The probability that Leah's bus is late on any day is 0.2. Whether it is late is independent from one day to the next. (a) Draw a tree diagram for Monday and Tuesday. (2 marks) (b) Work out the probability that the bus is late on both days. (2 marks) (c) Work out the probability that the bus is late on exactly one of the two days. (2 marks)
Show the model answer
(a) Monday: late 0.2, not late 0.8 (B1). From each Monday branch, Tuesday: late 0.2, not late 0.8 (B1). (b) 0.2 × 0.2 (M1) = 0.04 (A1) (c) 0.2 × 0.8 + 0.8 × 0.2 (M1) = 0.16 + 0.16 = 0.32 (A1)
Two picks without replacement
3 marks5
Draw the tree: the second pick is out of one fewer (here 10, then 9).
Reduce the numerator only for the colour already taken.
Find each path you need (here RR and YY), then add them.
Keep the same denominator (here 90) so the fractions are easy to add.
Example. A bag contains 4 red counters and 6 yellow counters. Tom takes a counter at random and does not put it back. He then takes a second counter at random. Work out the probability that the two counters are the same colour.
Show the model answer
P(RR) = \(\frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\) (M1 for either product) P(YY) = \(\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\) \(\frac{12}{90} + \frac{30}{90}\) (M1 for adding both products) \(= \frac{42}{90} = \frac{7}{15}\) (A1)
Shortcuts and memory tricks
'And' means multiply, 'or' means add.
Quick check: every pair of branches adds up to 1, and all the end results add up to 1.
Without replacement, the denominators count down: 10, then 9, then 8.
Leave fractions unsimplified until the end: with a common denominator (like 90) the paths are easy to add.
For 'at least one', 1 − P(none) is much quicker than adding every other path.
Where marks are lost
Adding along the branches instead of multiplying.
Forgetting that 'one of each' has two paths: red then yellow, and yellow then red.
Without replacement: not reducing the total for the second pick, or reducing the count of the wrong colour.
Adding the denominators when adding fractions: \(\frac{12}{90} + \frac{30}{90} = \frac{42}{90}\), not \(\frac{42}{180}\).
Writing branch probabilities that do not add up to 1, such as 0.3 and 0.6.
In a 'show that' question, jumping straight to the given equation without showing the lines in between.
Exam technique
Write a probability on every branch, even ones you think you will not need: completing the tree often earns marks on its own.
Write down each path you are adding, e.g. P(RR) + P(YY), so your method marks are clear even if you make a slip.
Look for the words 'replaces', 'does not replace' and 'eats': they tell you whether the events are independent.
Give exact answers (fractions, or exact decimals) where you can. If you have to round, keep at least 3 significant figures.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
On the way to work, Bella drives through two sets of traffic lights. The probability that the first set of lights is on red is 0.7. The probability that the second set of lights is on red is 0.8. The two sets of lights work independently. The probability tree diagram shows this information.
[object Object]
(a) Work out the probability that both sets of lights are on red.[2]
(b) Work out the probability that neither set of lights is on red.[2]
Show the answer and mark scheme
(a)Answer: \(0.56\)
M1 for \(0.7 \times 0.8\)
A1 for \(0.56\) oe
Worked solution: \(0.7 \times 0.8 = 0.56\)
(b)Answer: \(0.06\)
M1 for \(0.3 \times 0.2\)
A1 for \(0.06\) oe
Worked solution: \(0.3 \times 0.2 = 0.06\)
Question 2Medium3 marks
A fair six-sided dice is rolled twice. Dan says, 'The probability of getting at least one 6 is \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).'
(a) Explain why Dan's method is wrong.[1]
(b) Work out the correct probability of getting at least one 6.[2]
Show the answer and mark scheme
(a)Answer: A 6 on the first roll and a 6 on the second roll can both happen, so the events are not mutually exclusive: adding counts (6, 6) twice.
C1 for explaining that the two events are not mutually exclusive (both can happen), so the probabilities cannot simply be added / (6, 6) is counted twice
Worked solution: The outcome (6, 6) is included in both \(\frac{1}{6}\)s, so adding them counts it twice.
(b)Answer: \(\frac{11}{36}\)
M1 for \(1 - \left(\frac{5}{6}\right)^2\) or \(\frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6}\) or 11 outcomes out of 36 listed
A1 for \(\frac{11}{36}\) oe
Worked solution: P(no 6) \(= \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\), so P(at least one 6) \(= 1 - \frac{25}{36} = \frac{11}{36}\).