A two-stage tree diagram shows every outcome of two events in order. Multiply along each path, then add the paths that give the outcome you want. This is how IGCSE questions on repeated trials, two draws and two-day weather are organised.
It builds directly on listing mutually exclusive outcomes, because each set of branches must cover every outcome of that stage exactly once.
How is a tree built?
- Stage 1. Draw one branch for each outcome of the first event and label it with its probability.
- Stage 2. From the end of every stage 1 branch, draw a full set of branches for the second event.
- Check each split. The branches leaving one point must add to 1.
- Write the path probability at the end, found by multiplying along the path.
- Choose the paths you need, add their probabilities, and check that all end probabilities total 1.
Worked example
Adam catches the school bus on two consecutive mornings. Each morning the bus is on time with probability 0.8 and late with probability 0.2, independently of the other morning.
Find (a) P(on time both mornings), (b) P(exactly one late), (c) P(at least one late).
Step 1, draw the tree. First morning: O (0.8) or L (0.2). From each, the second morning: O (0.8) or L (0.2).
| Path | Working | Probability |
|---|---|---|
| O then O | 0.8 × 0.8 | 0.64 |
| O then L | 0.8 × 0.2 | 0.16 |
| L then O | 0.2 × 0.8 | 0.16 |
| L then L | 0.2 × 0.2 | 0.04 |
Step 2, check the total. 0.64 + 0.16 + 0.16 + 0.04 = 1.00, so the tree is consistent.
Step 3, (a). P(on time both mornings) = 0.64.
Step 4, (b). “Exactly one late” is O then L or L then O. Add: 0.16 + 0.16 = 0.32.
Step 5, (c). “At least one late” is everything except O then O. P = 1 − 0.64 = 0.36.
Check (c) a second way. Add the three paths containing a late bus: 0.16 + 0.16 + 0.04 = 0.36. Both routes agree.
The mistake to watch for
Probabilities are added along a path instead of multiplied.
Mistaken working: P(on time, then late) = 0.8 + 0.2 = 1.
A probability of 1 would mean it is certain, which cannot be right when a late bus is possible.
Correction. “On time first morning and late second morning” needs both to happen, so multiply: 0.8 × 0.2 = 0.16.
The mirror-image slip is multiplying between paths. For “exactly one late”, multiplying 0.16 × 0.16 gives 0.0256, which is far too small. Alternatives are added, steps are multiplied.
A useful sense check is that an “and” answer should be smaller than each individual probability, and an “or” answer should be larger than each path it includes.
Check yourself
Try these, then open each answer.
1. A biased coin has P(head) = 0.6 and P(tail) = 0.4. It is tossed twice. Find P(two heads) and P(one head and one tail).
Show answer
Two heads: 0.6 × 0.6 = 0.36. HT: 0.6 × 0.4 = 0.24 and TH: 0.4 × 0.6 = 0.24, so one of each is 0.24 + 0.24 = 0.48. Check with TT = 0.16: 0.36 + 0.48 + 0.16 = 1.
2. A bag has 2 red and 3 blue counters. A counter is taken, its colour noted and it is put back. Then a second counter is taken. Find P(both the same colour).
Show answer
Because it is replaced, each draw has P(red) = 2/5 and P(blue) = 3/5. Both red: 2/5 × 2/5 = 4/25. Both blue: 3/5 × 3/5 = 9/25. Add: 4/25 + 9/25 = 13/25.
3. The probability of passing a test is 0.7. Two separate tests are taken independently. Find P(pass at least one).
Show answer
The opposite is failing both: 0.3 × 0.3 = 0.09. So P(at least one pass) = 1 − 0.09 = 0.91. Check by paths: PP 0.49 + PF 0.21 + FP 0.21 = 0.91.
Where this leads next
The branch probabilities above stayed the same at stage 2 because the events were independent. Next, see when multiplication needs independence and then how the second stage changes when nothing is replaced. The probability tree and counting board lets you build a tree and see its total, and the non-calculator working trainer helps with the fractions.
If a tree is clear when you watch one being drawn but blank when you face an unfamiliar context, that is a good question to bring to online one-to-one Mathematics tuition.