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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Multi-stage experiments
Stages: Each branch level is one stage. For this spinner, each spin is independent: red stays 0.4 and blue stays 0.6 on the second spin.
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
Factory A supplies 60% of the items in a shop; factory B supplies 40%. Of A’s items, 2% are defective. Of B’s items, 3% are defective.
One item is chosen at random. Find the probability that it is defective.
Solution
- Defective from A: multiply along A → defective.
- Defective from B: multiply along B → defective.
- Either factory can supply the defective item, so add these two paths.
Final answer
P(defective) = 0.024 = 2.4%.
Completing a tree diagram
Branches from one point add to 1: Read left to right: first the weather, then whether the bus is late. At each split, the two probabilities add to 1.
First split: weather. Second split: bus late or on time. Follow both paths that end at Late.
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Fill in the missing branches
Use the same bus tree. Fill in these three missing branch probabilities:
(a) No rain.
(b) Bus on time if it rains.
(c) Bus on time if it does not rain.
Solution
- (a) Rain and no rain cover every day.
- (b) On rainy days, the bus is either late or on time.
- (c) On dry days, the bus is either late or on time.
Final answer
(a) 0.7 (b) 0.75 (c) 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.
Know your predicted grade
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Find the missing probability: two bags
What does b mean?: b is the chance of red from bag B only. The given 60% is the chance of red overall, after either bag is chosen. These are different probabilities.
There are two ways to pick red: bag A → red, or bag B → red. Add those paths.
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Bag B then red
(a) Write the probability of choosing bag B AND picking red, using b.
Solution
- Bag B has chance 0.5. Red from bag B has chance b. Multiply along this path.
Final answer
0.5b.
Part (b)
(b) Find b: the chance of red from bag B.
Solution
- Both red paths together give the known overall red chance of 0.6.
- Subtract the contribution from bag A.
- Divide by the chance of choosing bag B.
Final answer
b = 0.4: 40% of bag B’s counters are red.
Part (c)
(c) What is the chance of blue from bag B?
Solution
- Bag B contains only red and blue counters. Their probabilities add to 1.
Final answer
0.6, or 60%.
Exam tip: Write the equation before solving: setting it up earns the method mark even if the arithmetic slips.
AA exam practice: a quadratic probability
An unknown branch can appear in more than one factor. Write the probability equation, solve it, then check every root against the original branch probabilities. Work without a calculator.
Worked example [5 marks]
Two games are independent. The chance of winning the first is p; the chance of winning the second is p + 1/2. The chance of winning both is 3/16. Find p.
Step by step
- Multiply the two win probabilities.
- Multiply by 16 and rearrange.
- Factorise to find both roots.
- Both p and p + 1/2 must lie between 0 and 1. Reject the negative root.
Final answer
p = 1/4. The second win probability is 3/4.
Try the next question before opening its solution.
Your turn [5 marks]
Independent events A and B satisfy P(A) = x, P(B) = x + 1/3 and P(A ∩ B) = 2/9. Find x and justify rejecting the other root.
Step by step
- Use independence, then clear the denominators.
- Factorise.
- A probability cannot be negative. The valid probabilities are 1/3 and 2/3.
Final answer
x = 1/3; reject −2/3.
Exam tip: Show the equation and the rejected root. Do not give only the positive answer: explain why the other value is invalid.
Additional AA exam practice
IB-style question — with replacement
A bag has 3 red and 2 white balls.
A ball is drawn, replaced, then another is drawn.
Find the probability both are red.
Step by step
- P(red) is 3/5 each draw (replaced).
- Multiply along the red–red path.
Final answer
P(both red) = 9/25.
IB-style question — without replacement
From the same bag (3 red, 2 white), two balls are drawn without replacement.
Find the probability both are red.
Step by step
- First red 3/5; second red now 2/4.
- Multiply.
Final answer
P(both red) = 3/10.
IB-style question — algebraic tree (no replacement)
A bag contains x red counters and 4 white counters. Two counters are drawn without replacement. The probability that both are red is 1⁄3.
Find x.
Step by step
- Down the 'red then red' branch (the second draw has one fewer red and one fewer total).
- Cross-multiply and expand into a quadratic.
- Factor; reject the negative root (a count can't be negative).
Final answer
x = 6 red counters (check: (6⁄10)(5⁄9) = 1⁄3).
IB-style question — one of each colour
From the bag (3 red, 2 white), two are drawn without replacement.
Find the probability of one red and one white (in any order).
Step by step
- Two matching paths: red-then-white and white-then-red.
- Add the paths.
Final answer
P(one of each) = 3/5.
IB-style question — at least one
A box has 5 green and 3 yellow counters.
Two are taken without replacement.
Find the probability of at least one green.
Step by step
- Use the complement: 1 − P(no green) = 1 − P(both yellow).
- Multiply along the yellow–yellow path.
- Subtract from 1.
Final answer
P(at least one green) = 25/28.