GCSE · Probability

Tree Diagrams and Combined Events

What is the chance of two things both happening, like picking two red sweets, or a six on both dice? A tree diagram lays out every possible path so you can multiply along a path and add the paths you want. This page builds trees step by step: independent events, picking without putting back, "at least one", and three-stage trees.

GCSE subject content: record, describe and analyse the frequency of outcomes using frequency trees; calculate the probability of independent and dependent combined events, including using tree diagrams and other representations, and know the underlying assumptions

Before you start: what you need from KS3

1 · Frequency trees: counting first

Before we use probabilities, let's use counts. A frequency tree starts with a whole group, splits it into two parts, then splits each part again. Each box holds a frequency (how many). The one rule: the two boxes coming out of a box add up to that box. So whenever a split has one gap, you can fill it by subtracting.

Fill in the four empty boxes from the information given, then press Check. The tree updates as you type.

Once the tree is full, a probability is just a box divided by the box it came from at the start: "a person is picked at random from everyone" means divide by the very first number. Frequency trees are the bridge to probability trees: replace each count by a fraction of the box before it and you get the branches of section 3.

2 · Independent events

A combined event is two (or more) things happening together, like "spin red and roll a 6". Two events are independent if one happening does not change the probability of the other. A spinner can't see what the dice did, so they're independent. Events that do affect each other are dependent.

For independent events there's a short-cut: P(A and B) = P(A) × P(B). To see why it's a multiplication, draw every equally likely outcome as a grid: spinner sections across, dice faces down. Every cell is equally likely. The orange cells are "red and low enough": a block of (red columns) × (good rows) cells.

P(A) red
P(B) dice
P(A and B)

So the multiplication isn't a rule to memorise: it's counting cells in a rectangle. It only works because the spinner doesn't change the dice. Spotting whether events are independent is a skill in itself, and exam questions sometimes ask what you have assumed. Decide for each pair:

3 · Tree diagrams: the AND rule and the OR rule

A tree diagram shows a combined event in stages. From the start, draw one branch for each outcome of the first stage and write its probability on the branch. From the end of each branch, draw the branches for the second stage. Each route from the start to an end is a path, and each path is one combined outcome.

Two rules do everything:

Here a biased spinner lands on red (R) or not red (R′), and a biased coin lands heads (H) or tails (T). Set the probabilities, then tap the end results you want (or use a button) and watch the working.

Two checks that catch most mistakes. Every pair of branches from the same point adds up to 1 (something must happen). And all the end results add up to 1 (you must end up on exactly one path). If a probability ever comes out bigger than 1, you have added where you should have multiplied.

4 · Building a tree from a bag

A bag holds red and blue counters. You take one at random, note its colour, then take a second. There are two ways to do it:

Choose the bag, then press Next step to build the tree one stage at a time. Switch between with and without replacement and watch the second-stage branches change.

5 · Dependent events: the denominator drops

Here is "without replacement" happening one counter at a time. Take counters out and watch the fraction for the next pick. The bottom number always goes down by 1 (one fewer counter). The top number goes down by 1 only if you took that colour. Under the bag, the probability of the exact sequence you took is built up by multiplying.

Next pick: P(red)
Next pick: P(blue)

6 · "At least one" = 1 − P(none)

"At least one six" means one six, or two sixes, or more. On a tree that's almost every path, which is a lot of adding. But there is only one path with no six at all. Since all the paths add up to 1:

P(at least one) = 1 − P(none)

Roll a fair dice more times and watch the tree. S means a six (probability 1/6) and N means not a six (5/6). Orange paths have at least one six; only the grey one doesn't.

7 · Three-stage trees

Nothing new, just a bigger tree. Each path now has three branches to multiply, and with two outcomes at each stage there are 2 × 2 × 2 = 8 paths. Questions like "exactly two" are where trees earn their keep: you have to find every path with exactly two, and there are three of them. Pick an experiment, then tap paths or use a button.

8 · Spot the mistake

Each student below has made one of the classic tree-diagram mistakes. Look at their tree and working, and pick what went wrong.

Common mistakes and exam traps

Worked examples

P(A) = 0.6 and P(B) = 0.3, and A and B are independent. Find P(both), P(neither) and P(exactly one).
Complete the branches first: P(A′) = 1 − 0.6 = 0.4 and P(B′) = 1 − 0.3 = 0.7.
Both (A and B): multiply along: 0.6 × 0.3 = 0.18.
Neither (A′ and B′): 0.4 × 0.7 = 0.28.
Exactly one: two paths. A and B′: 0.6 × 0.7 = 0.42. A′ and B: 0.4 × 0.3 = 0.12. Add: 0.42 + 0.12 = 0.54.
Check: the four end results 0.18 + 0.42 + 0.12 + 0.28 = 1 ✓, and both + neither + exactly one = 0.18 + 0.28 + 0.54 = 1 ✓.
A bag has 5 red and 3 green sweets. Zara eats two at random. Find the probability that they are the same colour.
Spot the dependence: she eats the first sweet, so this is without replacement. 8 sweets, then 7.
Both red: 5/8 × 4/7 = 20/56. (One red has gone: 4 red out of 7.)
Both green: 3/8 × 2/7 = 6/56.
Same colour = both red OR both green: 20/56 + 6/56 = 26/56 = 13/28.
Check: different colours = 5/8 × 3/7 + 3/8 × 5/7 = 15/56 + 15/56 = 30/56, and 26/56 + 30/56 = 1 ✓.
A fair dice is rolled three times. Find the probability of at least one six.
Use the complement: at least one six = 1 − P(no sixes).
No sixes: not a six every time, and the rolls are independent: 5/6 × 5/6 × 5/6 = 125/216.
Answer: 1 − 125/216 = 216/216 − 125/216 = 91/216.
Why not 1/6 + 1/6 + 1/6 = 1/2? That counts rolls with two or three sixes more than once. With 6 rolls it would give 1, "certain", which is clearly false.
The probability that Ella wins a game of chess is 0.6. She plays three games, and the results are independent. Find the probability that she wins exactly two.
Branches: W = 0.6, L = 1 − 0.6 = 0.4 at every stage.
Paths with exactly two wins: WWL, WLW, LWW. Three paths.
Each path: 0.6 × 0.6 × 0.4 = 0.144 (the same three numbers in a different order, so the same product).
Add: 0.144 + 0.144 + 0.144 = 3 × 0.144 = 0.432.
200 students: 85 are boys, 40 of the boys have a school dinner, and 72 of the girls have a school dinner. Complete a frequency tree and find P(a student picked at random has a school dinner).
First split: girls = 200 − 85 = 115.
Boys split: 40 dinner, so 85 − 40 = 45 no dinner.
Girls split: 72 dinner, so 115 − 72 = 43 no dinner. Check: 40 + 45 + 72 + 43 = 200 ✓.
Probability: school dinner = 40 + 72 = 112 out of 200. P = 112/200 = 14/25.
Exam-style Higher: a bag has n sweets, 6 of them orange. Two are taken without replacement. P(both orange) = 1/3. Show that n² − n − 90 = 0 and find n.
Write the tree path: P(both orange) = 6/n × 5/(n − 1) = 30/(n(n − 1)).
Set it equal to 1/3: 30/(n(n − 1)) = 1/3, so n(n − 1) = 90.
Expand: n² − n = 90, so n² − n − 90 = 0.
Factorise: (n − 10)(n + 9) = 0, so n = 10 or n = −9. A number of sweets can't be negative, so n = 10.
Check: 6/10 × 5/9 = 30/90 = 1/3 ✓.

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