Tree diagram structure
How tree diagrams work: Branches show all outcomes.
Label each branch with probability.
Path probability: multiply along branches.
Multiply along the branches and add the end-paths; for conditional probability restrict to one branch first.
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Worked example
Bag: 3 red, 2 blue.
Draw 2 (no replace).
Show tree and find P(2 red).
Solution
- Branch 1: First red (3/5)
- Sub-branch: Second red given first red (2/4)
- Path prob: (3/5)×(2/4)=6/20=3/10
- Branch 2: First blue (2/5) leads to second outcomes
Final answer
P(RR)=3/10. Multiply along red path.
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Multi-stage experiments
Stages: Each branch level represents one stage.
Second stage branches depend on first outcome.
Worked example
Spinner spun twice: P(red)=0.4, P(blue)=0.6.
Find all outcomes and probabilities.
Solution
- RR: 0.4×0.4=0.16
- RB: 0.4×0.6=0.24
- BR: 0.6×0.4=0.24
- BB: 0.6×0.6=0.36
- Total: 0.16+0.24+0.24+0.36=1
Final answer
All paths shown, sum=1.
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Combining outcomes
Worked example
From defective example: Find P(defective).
Solution
- P(defective)=P(D|A)×P(A)+P(D|B)×P(B)
- =(0.02)×(0.6)+(0.03)×(0.4)
- =0.012+0.012=0.024
Final answer
P(defective)=0.024. Add paths leading to defective.
Completing a tree diagram
Branches from one point add to 1: Each split covers every outcome, so the missing branch is 1 minus the others. On the second stage, write the probability GIVEN the first branch.
Tap each step: first-stage branches, second-stage branches, multiply along a path, add the paths.
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Fill in the missing branches
The probability that it rains on a school day is 0.3. If it rains, the probability that a bus is late is 0.25. If it does not rain, the probability that the bus is late is 0.1. Write down the four missing probabilities on the tree (no rain; on time given rain; late and on time given no rain).
Solution
- No rain: 1 − 0.3 = 0.7.
- On time given rain: 1 − 0.25 = 0.75.
- Given no rain: late 0.1, on time 1 − 0.1 = 0.9.
Final answer
No rain 0.7; on time | rain 0.75; late | no rain 0.1; on time | no rain 0.9.
Independent events on a tree?
The second-stage branches are the SAME after every first branch.
Example: two switches, each fails with 0.05: both pairs of second branches are 0.05 and 0.95.
"At most one" from a tree?
Add every path with 0 or 1, or use 1 − P(both).
Example: two arrows, each hits with 0.6 → 1 − 0.6 × 0.6 = 0.64.
Exam tip: Write each branch value ON the diagram you are asked to complete: the marks are for the diagram.
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An unknown branch: an equation from the total
Add the paths, set equal to the total: When an overall probability is given, add every path that leads to it. The unknown branch is the only letter in the equation.
Tap Add and solve: the two "before 8:00" paths add to the given total, so b is the only unknown.
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Arriving before 8:00
40% of the students in a school cycle to school. Of those who cycle, 75% arrive before 8:00. Let b be the probability that a student who does not cycle arrives before 8:00. Overall, 54% of students arrive before 8:00.
(a) Write down an expression, in terms of b, for the probability that a student does not cycle and arrives before 8:00.
(b) Find b.
(c) Hence write down the probability that a student who does not cycle arrives at 8:00 or later.
Solution
- (a) Multiply along the "does not cycle" path: 1 − 0.4 = 0.6.
- (b) Add both "before 8:00" paths and set them equal to the overall value.
- Solve.
- (c) The other branch from the same point.
Final answer
(a) 0.6b. (b) b = 0.4. (c) 0.6.
Exam tip: Write the equation before solving: setting it up earns the method mark even if the arithmetic slips.