Tree Diagrams and Combined Events
A probability tree follows the stages of an experiment. It records the chance at each stage, then combines complete paths to answer questions about more than one event.
Foundation & Higher. Independent and dependent trees, including without replacement, are shared content. The final algebraic model is a Higher extension. Use Probability Basics and Rules and Sample Spaces and Two-Way Tables first. Review Calculating with Fractions for exact products and sums.
Build and read a probability tree
Draw branches for the possible first-stage outcomes. From each endpoint, draw branches for the next-stage outcomes. Label each branch with its probability in that situation, and label what happened at the end of each complete path.
At each fork, exhaustive mutually exclusive branches sum to 1. Do not add every branch on the whole page: different forks describe different situations.
Multiply along a path to find the probability of all its stages happening. Add different complete paths when any of those mutually exclusive routes would satisfy the question.
Independent events: later chances stay the same
Events are independent when knowing the result of one does not change the probability of the other. Independence is a modelling assumption to check, not something guaranteed just because there are two events.
Player A succeeds with probability 3/5; player B succeeds with probability 7/10. Assume their shots are independent. In the diagram S means success and M means miss; the first letter is A’s result.
Worked example — multiply one path
P(both succeed) = 3/5 × 7/10 = 21/50. Of the trials in which A succeeds, the same fraction 7/10 also have B succeed. That is why the two fractions multiply.
Worked example — add the two routes to exactly one success
A succeeds and B misses: 3/5 × 3/10 = 9/50. A misses and B succeeds: 2/5 × 7/10 = 14/50.
P(exactly one success) = 9/50 + 14/50 = 23/50. These are different complete paths, so they cannot both occur in the same pair of shots.
The remaining path, two misses, has probability 2/5 × 3/10 = 6/50. Check all four paths: (21 + 9 + 14 + 6)/50 = 1.
With replacement: restore the original probabilities
A bag contains 3 red and 2 blue counters. Draw a counter at random, replace it and mix thoroughly before drawing again. Replacement restores the composition, and the random mixed draws are modelled as independent.
Worked example — two draws with replacement
Every draw has P(red) = 3/5 and P(blue) = 2/5. P(two red) = 3/5 × 3/5 = 9/25.
For one of each colour, include both orders: P(RB or BR) = 3/5 × 2/5 + 2/5 × 3/5 = 12/25.
Do not multiply by 2 automatically in every problem. Here it works because the two path probabilities happen to be equal.
Without replacement: update each branch
Use the same bag of 3 red and 2 blue counters, but keep the first counter out. Only 4 counters remain for the second draw. If red was removed, 2 red remain; if blue was removed, all 3 red remain. The draws are dependent.
Worked example — two red, without replacement
P(RR) = 3/5 × 2/4 = 3/10. Both the total and the red count change after the first red draw.
Worked example — different colours, in either order
P(RB) = 3/5 × 2/4 = 3/10. P(BR) = 2/5 × 3/4 = 3/10. Their sum is 3/5.
P(BB) = 2/5 × 1/4 = 1/10. The four complete paths total 3/10 + 3/10 + 3/10 + 1/10 = 1.
Multiplication still works for dependent events, but use the second-stage probability from the correct branch. Multiplying unchanged original probabilities would give the wrong answer.
Dependence in a context
Dependence is not limited to counters. Suppose P(snow) = 0.2. The chance of a slow journey is 0.6 on a snowy day and 0.1 on a day without snow. The slow-journey chance changes with the first-stage situation.
| Route | Product | P |
|---|---|---|
| Snow, then slow | 0.2 × 0.6 | 0.12 |
| No snow, then slow | 0.8 × 0.1 | 0.08 |
P(slow journey) = 0.12 + 0.08 = 0.20. The two routes cover all slow journeys. Adding 0.6 + 0.1 would ignore how frequently the two types of day occur.
At least one: sometimes use the complement
“At least one” includes one, two or more. Its opposite is “none”, which may be only one path. Use the complement when it avoids a long list.
Worked example — three independent trials
Each of three independent attempts succeeds with probability 0.4. P(no successes) = 0.6 × 0.6 × 0.6 = 0.216.
P(at least one success) = 1 − 0.216 = 0.784. This is different from exactly one success, which would require the three separate SMM, MSM and MMS paths.
Higher extension — an unknown number of counters
A probability model can give an equation. Define the unknown, build the correct probability expression, solve it, then reject solutions that cannot describe the original objects.
Worked example — use a probability to find a bag’s contents
A bag has 3 red and n blue counters. Two are drawn without replacement, and P(two red) = 1/5. Therefore (3/(n + 3)) × (2/(n + 2)) = 1/5.
Cross-multiply: (n + 3)(n + 2) = 30. Expanding and rearranging gives n² + 5n − 24 = 0, so (n + 8)(n − 3) = 0.
The roots are −8 and 3. A counter count cannot be negative, so n = 3. Check: 3/6 × 2/5 = 1/5. For the algebraic method, review Quadratic Equations.
Think Like a Mathematician
Mutually exclusive means “cannot both happen in one trial”; independent means “one result does not change the other’s chance”. On one die roll, “1” and “6” are mutually exclusive, but knowing the result is 1 makes the chance of 6 zero. They are not independent.
Common Mistakes
Adding down a path: a complete sequence uses multiplication of the appropriate branch probabilities.
Ignoring another valid order: one red and one blue usually includes RB and BR.
Keeping the old denominator: without replacement, fewer objects remain; also update the numerator for the removed colour.
Multiplying original probabilities for dependent events: use the chance appropriate to what has already happened.
Practice
1. At one fork, the only two branches have probabilities 0.65 and p. Find p.
The branches total 1, so p = 1 − 0.65 = 0.35.
2. Independent events A and B have probabilities 0.3 and 0.8. Find P(A and B).
0.3 × 0.8 = 0.24.
3. Two independent attempts have success probabilities 0.4 and 0.75. Find P(exactly one success).
Success then failure: 0.4 × 0.25 = 0.10. Failure then success: 0.6 × 0.75 = 0.45. Add the routes: 0.55.
4. A bag has 4 red and 3 blue counters. Draw, replace and mix, then draw again. Find P(red then blue).
4/7 × 3/7 = 12/49. Replacement restores all 7 counters.
5. Use that bag of 4 red and 3 blue without replacement. Find P(two red).
4/7 × 3/6 = 12/42 = 2/7.
6. Without replacement from that bag, find P(one of each colour).
RB: 4/7 × 3/6 = 2/7. BR: 3/7 × 4/6 = 2/7. Add: 4/7.
7. P(rain)=0.3. P(late on a rainy day)=0.8 and P(late on a dry day)=0.1. Find P(late).
Rain and late: 0.3 × 0.8 = 0.24. Dry and late: 0.7 × 0.1 = 0.07. Total = 0.31.
8. Three independent attempts each succeed with probability 0.2. Find P(at least one success).
P(none) = 0.8³ = 0.512. The complement is 1 − 0.512 = 0.488.
9. Explain why “mutually exclusive” and “independent” do not mean the same thing.
Mutually exclusive events cannot happen together. Independent events leave each other’s probabilities unchanged. Two mutually exclusive events with positive probabilities cannot be independent: knowing one happened rules the other out.
10. Higher: a bag has 4 red and n blue counters. Two are drawn without replacement. P(two red)=2/15. Find n.
(4/(n+4)) × (3/(n+3)) = 2/15, so (n+4)(n+3)=90. Then n²+7n−78=0, giving (n+13)(n−6)=0. Reject −13; n=6. Check: 4/10 × 3/9 = 2/15.
Frequently Asked Questions
Are without-replacement questions Higher-only?
No. Independent and dependent combined events, including suitable without-replacement trees, are part of the shared GCSE curriculum. More demanding algebraic models and conditional inference add Higher difficulty.
Does “AND multiply, OR add” always work unchanged?
Use the right conditions. Multiplication along a tree uses the branch probability appropriate to earlier results. Adding requires disjoint routes, or an overlap correction. The words alone do not justify multiplying unchanged probabilities or double-counting overlaps.