Revision notes for Edexcel GCSE Maths Probability Trees. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Maths (1MA1) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for Edexcel GCSE Maths Probability Trees. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Maths (1MA1) specification, so the content matches what's examinable rather than general Maths background.
A probability tells you how likely something is. It can be written as a fraction, decimal, or percentage. A probability of 0 means impossible, and a probability of 1 means certain.
Event
An event is an outcome, or a group of outcomes, that you are interested in. For example, choosing a blue counter or winning a match.
Complement
The complement of an event means “not that event”. An event and its complement always add to 1, so P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)P(not A)=1−P(A).
Finding a missing branch label
In a game, the probability of winning is 0.65. What probability should go on the “not win” branch?

The two outcomes are “win” and “not win”, so their probabilities must add to 1.
Subtract the win probability from 1:
P(not win)=1−0.65=0.35P(\text{not win}) = 1 - 0.65 = 0.35P(not win)=1−0.65=0.35Label the second branch 0.35.
A probability tree shows events happening in stages. For example, first choose from Bag A, then choose from Bag B.
Probability tree
A probability tree is a diagram with branches that show the possible outcomes of two or more events, with probabilities written on the branches.
At each split in the tree, the probabilities should add to 1.
Independent events
Two events are independent if the result of the first event does not change the probabilities for the second event.
For separate bags, separate games, or different days with the same chance each day, the second set of branches usually has the same labels every time.
Labelling a tree from two bags
Sofia has one bag with 2 red counters and 6 blue counters. A second bag has 4 red counters and 1 blue counter. She takes one counter from each bag.

For the first bag, there are 8 counters in total:
P(red from first bag)=28=14,P(blue from first bag)=68=34P(\text{red from first bag}) = \frac{2}{8} = \frac{1}{4}, \quad P(\text{blue from first bag}) = \frac{6}{8} = \frac{3}{4}P(red from first bag)=82=41,P(blue from first bag)=86=43For the second bag, there are 5 counters in total:
P(red from second bag)=45,P(blue from second bag)=15P(\text{red from second bag}) = \frac{4}{5}, \quad P(\text{blue from second bag}) = \frac{1}{5}P(red from second bag)=54,P(blue from second bag)=51Draw the first split for the first bag: red with 14\frac{1}{4}41 and blue with 34\frac{3}{4}43.
From the end of both first-bag branches, draw the second-bag branches: red with 45\frac{4}{5}54 and blue with 15\frac{1}{5}51.
Check each split adds to 1.
When probabilities change
If you take two items from the same bag without replacing the first item, the second probabilities can change. The examples here use separate bags, games, or days, so the probabilities repeat.
When a question asks for two things to happen together, such as “win both games” or “choose two blue counters”, you follow one complete path through the tree.
Route
A route is one complete path from the start of a probability tree to an end outcome.
AND means multiply
To find the probability of one complete route, multiply the branch probabilities along that route.
Winning two matches
Amira plays one badminton match and one table tennis match. The probability she wins badminton is 35\frac{3}{5}53, and the probability she wins table tennis is 710\frac{7}{10}107. Find the probability she wins both.

The “not win” branches would be 25\frac{2}{5}52 for badminton and 310\frac{3}{10}103 for table tennis.
“Wins both” means: win badminton and win table tennis.
Multiply the two win probabilities:
P(both wins)=35×710=2150P(\text{both wins}) = \frac{3}{5} \times \frac{7}{10} = \frac{21}{50}P(both wins)=53×107=5021The probability she wins both is 2150\frac{21}{50}5021.
Adding for ‘both’
Do not add probabilities when the question asks for both events. Adding would mean “one or the other”, not one full route.
Sometimes the result you want can happen in more than one way. For example, “different coloured ties” could mean black then red, or red then black.
OR means add routes
If a result can happen by more than one route, multiply along each route first, then add the route probabilities.
Different colours on two days
Eli wears either a navy tie or a grey tie. On any day, P(navy)=59P(\text{navy}) = \frac{5}{9}P(navy)=95. Find the probability he wears different colours on Monday and Tuesday.

First find the grey probability:
P(grey)=1−59=49P(\text{grey}) = 1 - \frac{5}{9} = \frac{4}{9}P(grey)=1−95=94Different colours can happen in two routes: navy then grey, or grey then navy.
Multiply along each route:
P(navy then grey)=59×49=2081P(grey then navy)=49×59=2081\begin{aligned} P(\text{navy then grey}) &= \frac{5}{9} \times \frac{4}{9} = \frac{20}{81} \\ P(\text{grey then navy}) &= \frac{4}{9} \times \frac{5}{9} = \frac{20}{81} \end{aligned}P(navy then grey)P(grey then navy)=95×94=8120=94×95=8120Add the two route probabilities:
P(different colours)=2081+2081=4081P(\text{different colours}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}P(different colours)=8120+8120=8140The probability he wears different colours is 4081\frac{40}{81}8140.
At least one
“At least one” means one or more. For two events, it includes first only, second only, and both.
You could list every successful route, but there is often a quicker method.
Shortcut for at least one
“At least one” is the opposite of “none”, so use P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})P(at least one)=1−P(none).
Late on at least one day
Sam catches a train on Saturday and Sunday. The probability the train is late on any day is 0.3. Find the probability it is late on at least one of the two days.

The probability the train is not late is 0.7.
“None late” means not late on Saturday and not late on Sunday:
P(none late)=0.7×0.7=0.49P(\text{none late}) = 0.7 \times 0.7 = 0.49P(none late)=0.7×0.7=0.49“At least one late” is the opposite of “none late”:
P(at least one late)=1−0.49=0.51P(\text{at least one late}) = 1 - 0.49 = 0.51P(at least one late)=1−0.49=0.51The probability it is late on at least one day is 0.51.
Not every event has only two outcomes. A game might have win, draw, or lose. The important rule is the same: all branch probabilities from the same point must add to 1.
Win, draw or lose
A player plays two rounds of a game. The probability of a win is 0.5, and the probability of a draw is 0.2. Find the probability the player wins both rounds.

Find the missing lose probability:
P(lose)=1−0.5−0.2=0.3P(\text{lose}) = 1 - 0.5 - 0.2 = 0.3P(lose)=1−0.5−0.2=0.3For each round, the tree has three branches: win 0.5, draw 0.2, lose 0.3.
Because the probabilities are the same for each round, draw the same three branches after each first-round outcome.
“Wins both rounds” is the win then win route:
P(win both)=0.5×0.5=0.25P(\text{win both}) = 0.5 \times 0.5 = 0.25P(win both)=0.5×0.5=0.25The probability the player wins both rounds is 0.25.
In the exam
Write the missing branch probabilities first, and check every split adds to 1.
For “both” or “and”, follow one route and multiply along it.
For “or”, “different”, or “at least one”, find all suitable routes and add them, or use the complement shortcut.
Check yourself
Can you label a two-stage tree when given counts in two bags?
Can you explain why “navy then grey” and “grey then navy” are different routes?
When a question says “at least one”, what is the shortcut using “none”?
Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.
Test yourself on this topic, or move on to the next guide.
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