A tree diagram shows one event after another. Multiply the probabilities along each path to reach an outcome, then add the paths that give the event you want.
Probabilities go beside the branches and outcomes at the ends. Multiply ALONG a path for 'this and then that'; ADD the results of different paths for 'this or that'. The branches from any one point add up to 1.
Fig. 1: two counters taken with replacement from a bag of 3 red and 2 blue; the highlighted paths give one of each colour, 6/25 + 6/25 = 12/25
Section 01
Drawing the tree.
In the lab, set up a bag of counters and build the tree one pick at a time. Choose an event and watch which paths light up: multiply along each lit path, then add them.
Worked example With replacement
A bag holds 3 red and 2 blue counters. A counter is taken, its colour noted, and it is put back. A second counter is then taken. Find (a) P(both red) (b) P(one of each colour).
Put back, so the second pick has the same probabilities: \(\tfrac{3}{5}\) red, \(\tfrac{2}{5}\) blue (Fig. 1).
(a) Multiply along R then R: \(\tfrac{3}{5} \times \tfrac{3}{5} = \tfrac{9}{25}\).
(b) Two paths: RB gives \(\tfrac{3}{5} \times \tfrac{2}{5} = \tfrac{6}{25}\), and BR gives \(\tfrac{2}{5} \times \tfrac{3}{5} = \tfrac{6}{25}\).
Add the paths: \(\tfrac{6}{25} + \tfrac{6}{25} = \tfrac{12}{25}\).
Check by adding: \(\tfrac{6}{25} + \tfrac{6}{25} + \tfrac{4}{25} = \tfrac{16}{25}\).
Answer: \(\tfrac{16}{25}\)
Examiner trap
Forgetting one of the two orders. 'One red and one blue' can happen as RB OR as BR, so there are two paths to add.
Section 02
Independent events.
Trees work for any events that follow each other, not just counters. If one event does not affect the other, the second set of branches repeats the same probabilities.
Worked example Two tests
The probability that Mia passes her theory test is 0.8 and her practical test is 0.6, independently. Find the probability that she passes (a) both (b) exactly one.
(b) Pass then fail: \(0.8 \times 0.4 = 0.32\). Fail then pass: \(0.2 \times 0.6 = 0.12\).
Add: \(0.32 + 0.12 = 0.44\).
Answer: (a) 0.48 (b) 0.44
Worked example Three throws
A fair coin is thrown three times. Find the probability of exactly two heads.
Each path of three throws has probability \(\tfrac{1}{2} \times \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{8}\).
Paths with exactly two heads: HHT, HTH, THH, which is 3 paths.
\(3 \times \tfrac{1}{8} = \tfrac{3}{8}\).
Answer: \(\tfrac{3}{8}\)
Your turn
The probability that it rains on any day is 0.3. Find the probability that it rains on exactly one of two days.
Rain then dry \(0.3 \times 0.7 = 0.21\); dry then rain \(0.7 \times 0.3 = 0.21\). Total \(0.42\).
Examiner trap
Adding along a path instead of multiplying gives nonsense: \(0.8 + 0.6 = 1.4\) is bigger than 1. Along a path, always multiply.
Stage 1 · Learn
Check your understanding
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On a tree diagram, to find the probability of one path you...
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What do the probabilities on the branches from any one point add up to?
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A fair coin is thrown twice. What is the probability of two heads?
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(Extended) A bag has 4 red and 6 blue counters. A red is taken and not replaced. What is the probability that the next counter is red?
Section 03
Without replacement.
(Extended) If the first item is NOT put back, the second pick has one fewer item in the bag, and one fewer of the colour already taken. The second-stage branches change.
Worked example Two green (Extended)
A bag holds 5 red and 3 green sweets. Two are taken without replacement. Find P(both green).
First pick: \(\text{P(G)} = \tfrac{3}{8}\).
Now 7 sweets are left, 2 of them green: \(\text{P(G second)} = \tfrac{2}{7}\).
(Extended) For the same bag, find the probability that at least one sweet is red.
The only other outcome is both green, so \(1 - \tfrac{3}{28} = \tfrac{25}{28}\).
Examiner trap
Without replacement, BOTH the numerator and the denominator can change. After a green is taken, green becomes \(\tfrac{2}{7}\) but red becomes \(\tfrac{5}{7}\), not \(\tfrac{5}{8}\).
Stage 2 · Exam
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The probability that a bus is full is 0.2, independently each day. Find the probability that it is full on exactly one of two days.
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A bag has 4 red and 6 blue counters. Two are taken WITH replacement. Find the probability that at least one is red.
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(Extended) A box has 7 red and 3 yellow beads. Two are taken without replacement. Find the probability that they are the same colour.
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A biased coin has P(head) = 0.6. It is thrown three times. Find the probability of exactly two heads.
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