What is a tree diagram?
A tree diagram shows all the possible outcomes of two or more events, one after another, as a set of branches. Each branch is labelled with a probability, and the diagram makes it easy to calculate the probability of combined outcomes.
The two golden rules:
- Multiply along the branches to find the probability of a particular sequence of outcomes (this represents "and")
- Add the end probabilities to find the probability of one outcome or another (this represents "or")
- The probabilities on branches from the same point must always add up to 1
- For independent events (e.g. picking with replacement), the probabilities on the second set of branches stay the same. For dependent events (e.g. picking without replacement), the second set of branches changes depending on the first outcome
A fair coin tossed twice. Multiply along each pair of branches to get the four end probabilities — they add up to 1 (¼ + ¼ + ¼ + ¼).
Independent events (with replacement)
Worked Example 1
A bag contains 3 red and 2 blue counters. A counter is picked, replaced, then a second counter is picked. Find the probability that both counters are red.
1
P(red) = 3/5 on each pick, since the counter is replaced — the probabilities don't change
2
Multiply along the branches: P(red, red) = 3/5 × 3/5
3
= 9/25
Answer9/25
Combining branches with "or"
Worked Example 2
Using the same bag (3 red, 2 blue, with replacement), find the probability of picking one counter of each colour.
1
"One of each colour" can happen two ways: red then blue, OR blue then red
2
P(red, blue) = 3/5 × 2/5 = 6/25. P(blue, red) = 2/5 × 3/5 = 6/25
3
Add the two paths: 6/25 + 6/25 = 12/25
Answer12/25
Dependent events (without replacement)
Worked Example 3
A bag contains 3 red and 2 blue counters. Two counters are picked without replacement. Find the probability that both are red.
1
First pick: P(red) = 3/5
2
After removing one red counter, there are 2 red and 2 blue left out of 4 total: P(red second) = 2/4
3
Multiply along the branches: 3/5 × 2/4 = 3/10
Answer3/10
Worked Example 4 — "At least one"
Using the same bag (3 red, 2 blue, without replacement), find the probability of picking at least one blue counter.
1
"At least one blue" is the opposite of "no blue at all" (i.e. both red)
2
From Worked Example 3, P(both red) = 3/10
3
P(at least one blue) = 1 − P(both red) = 1 − 3/10 = 7/10
Answer7/10
Practice questions
Work through each question before checking the answers.
Foundation (Grade 3–5)
Q1A spinner has two outcomes: Win (P = 0.3) and Lose (P = 0.7). It is spun twice. Find P(Win then Lose).Foundation
Q2A bag contains 4 red and 6 blue counters. A counter is drawn, replaced, then a second is drawn. Find P(both blue).Foundation
Q3On a tree diagram, the first set of branches has probabilities x and 0.65. Find the value of x.Foundation
Q4A biased coin has P(heads) = 0.7. It is flipped twice. Find P(two tails).Foundation
Q5On a tree diagram, the first branch has P(A) = 0.4. From A, the next branch has P(B) = 0.5. Find P(A then B).Foundation
Higher (Grade 5–7)
Q6A bag contains 5 red and 3 blue counters. Two counters are drawn with replacement. Find P(one of each colour).Higher
Q7A biased dice has P(rolling a 6) = 0.2. It is rolled twice. Find P(at least one 6).Higher
Q8A box contains 4 red pens and 6 blue pens. Two pens are taken without replacement. Find P(both red).Higher
Q9Using the box from Q8 (4 red, 6 blue, without replacement), find P(one pen of each colour).Higher
Q10A fair coin is tossed three times. Find P(exactly two heads).Higher
Higher — Hard (Grade 8–9)
Q11A bag contains 7 red and 3 blue counters. Two are drawn without replacement. Find P(at least one blue).Grade 8–9
Q12A bag contains n red counters and 5 blue counters. Two are drawn without replacement. If P(both blue) = 2/9, find n.Grade 8–9
Q13A biased coin has P(heads) = p. It is flipped twice. If P(two heads) = 0.49, find p, then find P(exactly one head).Grade 8–9
Q14A bag contains only red and blue counters, 12 in total. Two are drawn without replacement. P(both red) = 5/22. Find the number of red counters.Grade 8–9
Q15Explain, using the idea of branch probabilities changing between the first and second pick, why events "without replacement" are dependent while events "with replacement" are independent.Grade 8–9
Answers
Foundation (Q1–Q5)
Q10.21(0.3 × 0.7)
Q20.36(0.6 × 0.6 — replacement keeps P(blue) the same)
Q3x = 0.35(1 − 0.65)
Q40.09(0.3 × 0.3)
Q50.2(0.4 × 0.5)
Higher (Q6–Q10)
Q615/32(2 × (5/8 × 3/8) = 2 × 15/64)
Q70.36(1 − 0.8² = 1 − 0.64)
Q82/15(4/10 × 3/9 = 12/90)
Q98/15(P(RB) + P(BR) = 24/90 + 24/90 = 48/90)
Q103/8(3 arrangements of HHT × (1/2)³ = 3/8)
Higher — Hard (Q11–Q15)
Q118/15(1 − P(RR) = 1 − 42/90 = 1 − 7/15)
Q12n = 5(5/(n+5) × 4/(n+4) = 2/9 → (n+5)(n+4) = 90)
Q13p = 0.7; P(exactly one head) = 0.42(p² = 0.49 → p = 0.7; 2 × 0.7 × 0.3)
Q146 red counters(r(r−1)/132 = 5/22 → r² − r − 30 = 0 → r = 6)
Q15With replacement, the counter is put back before the second pick, so the total and the number of each colour stay the same — the second-branch probabilities are identical each time, making the events independent. Without replacement, removing a counter changes both the total and the count of that colour, so the second-branch probabilities depend on the first outcome — making the events dependent.
Common mistakes
Common Mistake 1
Adding along branches instead of multiplying
To find the probability of a specific sequence of outcomes (following one path through the tree), you must multiply the branch probabilities, not add them. P(red, then blue) = 3/5 × 2/5, not 3/5 + 2/5.
Common Mistake 2
Forgetting to change probabilities for "without replacement"
When an item isn't replaced, both the total and the count of the relevant outcome go down by one for the second pick. A common error is keeping the same fraction (e.g. still using 3/5) for the second branch instead of recalculating it.
Common Mistake 3
Missing a valid path when combining branches
"One of each colour" or "at least one" often includes more than one path through the tree. Before adding, check the diagram carefully for every route that satisfies the condition — it's easy to spot red-then-blue but forget blue-then-red.
Common Mistake 4
Not simplifying the final fraction
Tree diagram answers often come out as fractions with large denominators (e.g. 24/90). Always check whether the fraction can be simplified — 24/90 simplifies to 4/15 — as exam mark schemes usually expect the simplest form.
Exam tips
💡 Exam Tip 1
Draw the tree even if it isn't given
Even when a question doesn't provide a diagram, sketching one yourself makes it far easier to see every branch and avoid missing a path — and it earns method marks even if your final answer is wrong.
💡 Exam Tip 2
Label every branch as you draw it
Write the probability on every single branch, including ones you might think are "obvious". This makes it much less likely you'll accidentally use the wrong number when multiplying along a path.
💡 Exam Tip 3
Use "1 minus" for "at least one" questions
"At least one" is almost always easier to find using P(at least one) = 1 − P(none), rather than adding up every possible way of getting one, two, three, or more successes.
💡 Exam Tip 4
Check your branches sum to 1 at every stage
At each point on the tree, the branches leaving that point should add up to 1 — check this at the first set of branches and at each second set of branches, since with dependent events every set can be different.
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