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.
Interactive diagram
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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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Conditional probability from trees
Reading conditional prob: P(B|A) is the probability on second-stage branch GIVEN first stage was A.
Worked example
Factory: Machine A makes 60% of items, 2% defective.
Machine B makes 40%, 3% defective.
Find P(defective|Machine A).
Solution
- Tree: First stage: A (0.6) or B (0.4)
- From A: defective (0.02) or good (0.98)
- P(defective|A)=0.02 (second-level branch)
Final answer
0.02. Branch probability in conditional state.
IB-style question — reverse conditional [6 marks]
A shop is supplied bread by two bakeries. Bakery X supplies 70% of the loaves and Bakery Y supplies the other 30%. 4% of Bakery X's loaves are stale, while 10% of Bakery Y's loaves are stale. A loaf is chosen at random.
(a) Find the probability that the loaf is stale.
(b) Given that the chosen loaf is stale, find the probability that it came from Bakery Y.
Step by step
- (a) A loaf is stale along two tree paths: from X, or from Y. Find each path then add.
- Substitute the branch probabilities.
- (b) Use the conditional rule. The intersection 'Y and stale' is the Bakery-Y path from part (a).
- Evaluate.
Final answer
(a) P(stale) = 0.058. (b) P(Y | stale) = 0.030/0.058 = 0.517 (3 s.f.).
GDC walkthrough
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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.