Tree Diagrams and Combined Events
A tree diagram lays out every possible outcome of two or more events so you can’t miss one. There are exactly two operations, and nearly every lost mark comes from using the wrong one.
1. The two rules
MULTIPLY along the branches (left to right) — this gives the probability of one complete path. ADD between the paths (down the diagram) — when more than one path satisfies the question.
A way to remember it: moving along a path means “this and then this” → multiply. Choosing between paths means “this path or that path” → add.
Adding along the branches instead of multiplying is the most common error in this topic, and it was recorded again and again. The giveaway is an answer bigger than 1, or bigger than either individual probability — combining two events makes the outcome less likely, not more.
Two structural checks:
Each PAIR of branches must add to 1. If P(red) = 3/8, the other branch is 5/8. All the FINAL outcomes must add to 1. This checks the whole diagram in one line and is worth doing every time.
2. Drawing the diagram
For two events with two outcomes each, you get four paths.
Example: a bag with 3 red and 5 blue counters (8 total), two picked WITH replacement.
- First set of branches: 3/8 red, 5/8 blue
- Second set: identical, 3/8 and 5/8, because the counter went back
| Path | Calculation | Probability |
|---|---|---|
| Red, Red | 3/8 × 3/8 | 9/64 |
| Red, Blue | 3/8 × 5/8 | 15/64 |
| Blue, Red | 5/8 × 3/8 | 15/64 |
| Blue, Blue | 5/8 × 5/8 | 25/64 |
Check: 9 + 15 + 15 + 25 = 64, so the four add to 64/64 = 1 ✓
Label every branch with its outcome and its probability, and write the final probability at the end of each path. Marks are available for a correctly labelled diagram even before any calculation.
3. Without replacement
The second set of branches changes — both the numerator and the denominator drop.
Same bag, but the first counter is not returned:
- First branches: 3/8 red, 5/8 blue
- If red was taken first: 2 red and 5 blue remain → second branches are 2/7 and 5/7
- If blue was taken first: 3 red and 4 blue remain → second branches are 3/7 and 4/7
| Path | Calculation | Probability |
|---|---|---|
| Red, Red | 3/8 × 2/7 | 6/56 |
| Red, Blue | 3/8 × 5/7 | 15/56 |
| Blue, Red | 5/8 × 3/7 | 15/56 |
| Blue, Blue | 5/8 × 4/7 | 20/56 |
Check: 6 + 15 + 15 + 20 = 56/56 = 1 ✓
The denominator drops on every second branch, because one item has gone. Only the numerator that matches the first pick reduces. Forgetting the denominator was a specific recorded error.
The second set of branches is no longer identical — that is the whole difference from a “with replacement” tree, and it is why you must draw the diagram rather than reuse the first probabilities.
4. Answering the question
“Both red” — one path: 6/56 = 3/28
“One of each colour” — two paths, so add:
- 15/56 + 15/56 = 30/56 = 15/28
“One of each” always means two paths. Missing the second order halves the answer, and it was flagged repeatedly.
“At least one red” — use the complement:
- P(no red) = P(blue, blue) = 20/56
- P(at least one red) = 1 − 20/56 = 36/56 = 9/14
1 − P(none) is far quicker than adding three paths, and much less error-prone. “At least one” almost always signals this method.
“At least one” is not “exactly one”. At least one includes the both-red path; exactly one does not.
5. Conditional probability
A conditional probability is the probability of something given that something else has already happened.
On a tree diagram, the second set of branches is already conditional — that is exactly what “without replacement” encodes.
“Given that the first counter was red, what is the probability the second is blue?”
- Read straight off the second branch: 5/7
- You do not multiply — the first event is already known to have happened
If the question says “given that”, read the branch — don’t multiply along the path. Multiplying answers a different question (the probability of both), and this confusion was recorded in lessons.
6. Three events, and unknowns
Three events give eight paths. The rules don’t change: multiply along, add between. Just be systematic and check everything sums to 1.
Questions with an unknown: if a bag contains n counters and you’re told P(both red) = some value, form an equation from the product along that path and solve — often producing a quadratic.
7. Scope note
Some lessons covered permutations and combinations — committee-selection problems, code-counting with restricted digits, and nCr / nPr notation.
These are not on the 0580 syllabus — they belong to Additional Mathematics (0606) and A-level. If you are sitting 0580 only, tree diagrams and the listing methods above are all you need. Check your own syllabus document before revising nCr.
8. Mistakes that cost marks
Adding along the branches instead of multiplying.
Multiplying between paths instead of adding.
Not changing the second branches in a without-replacement question.
Forgetting to reduce the denominator.
Giving only one path for “one of each”.
Treating “at least one” as “exactly one”.
Multiplying when the question said “given that”.
Branches that don’t add to 1.
Leaving the diagram unlabelled.
Simplifying wrongly at the last step.
Frequently asked questions
When do I multiply on a tree diagram? Along the branches — for a single complete path.
When do I add? Between paths — when several outcomes satisfy the question.
How do I check my tree diagram? Each pair of branches adds to 1, and all the final probabilities add to 1.
What changes without replacement? The second set of branches — both the relevant numerator and the total decrease by one.
How do I find “one of each”? Add the two paths — one for each order.
How do I find “at least one”? 1 − P(none).
What is conditional probability? The probability of an event given another has occurred — read directly off the second branch.
Do I need nCr for 0580? No — that is 0606 material.
Quick revision checklist
- I multiply along branches and add between paths
- I check each pair of branches adds to 1
- I check all final outcomes add to 1
- I label every branch with outcome and probability
- I can draw a with replacement tree
- I can draw a without replacement tree, changing both parts of the fraction
- I know the two second-branch sets differ depending on the first pick
- I add both orders for “one of each”
- I use 1 − P(none) for “at least one”
- I know “at least one” ≠ “exactly one”
- I read conditional probabilities off the branch
- I can extend to three events
- I know nCr is not on 0580
These notes cover tree diagrams and combined events in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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