All subjects · Mathematics · WST01 (S1)

Tree diagrams, with and without replacement

Tree diagrams organise two or more stages of an experiment. Multiply along the branches and add between the paths you want.

Drawing a tree

Each set of branches covers every outcome at that stage, so the probabilities on them add to 1. Later branches show conditional probabilities: the probability of the second outcome given the first.

With and without replacement

A bag has 5 red and 3 blue counters. Two are taken.

With replacement: the second draw has the same probabilities, 5/8 and 3/8.
Without replacement: after a red is taken there are 4 red and 3 blue left, so P(red second | red first) = 4/7.

First	Second	Probability (without replacement)
Red 5/8	Red 4/7	20/56
Red 5/8	Blue 3/7	15/56
Blue 3/8	Red 5/7	15/56
Blue 3/8	Blue 2/7	6/56

Using the tree

Multiply along a path for 'this and then that'.
Add the paths that satisfy the event.
'At least one' is usually quickest as 1 - P(none).
Conditional questions: P(first red | same colour) = P(red, red) / P(same colour). Show the working: examiners give no marks for a bare answer to a conditional probability.

Worked example

Without replacement, find the probability that the two counters are different colours.

P(red, blue) = 5/8 × 3/7 = 15/56
P(blue, red) = 3/8 × 5/7 = 15/56
Add the two paths.

Answer: 30/56 = 15/28

Key idea: Multiply along branches, add between paths; without replacement, change the second-stage fractions.

Check yourself

5 red and 3 blue counters; two are taken without replacement. Find P(two reds). Give your answer as a fraction in its simplest form.
Show answer
5/14. 5/8 × 4/7 = 20/56 = 5/14.
5 red and 3 blue counters; two are taken with replacement. Find P(two reds). Give your answer as a fraction in its simplest form.
Show answer
25/64. 5/8 × 5/8 = 25/64.
5 red and 3 blue counters; two are taken without replacement. Find P(different colours). Give your answer as a fraction in its simplest form.
Show answer
15/28. 15/56 + 15/56 = 30/56 = 15/28.
5 red and 3 blue counters; two are taken without replacement. Find P(at least one blue). Give your answer as a fraction in its simplest form.
Show answer
9/14. 1 - P(two reds) = 1 - 5/14 = 9/14.
5 red and 3 blue counters; two are taken without replacement. Given that the counters are the same colour, find the probability that they are both red. Give your answer as a fraction in its simplest form.
Show answer
10/13. P(same) = 20/56 + 6/56 = 26/56. P(RR | same) = (20/56)/(26/56) = 10/13.
P(rain) = 0.3. If it rains, P(late) = 0.4; if not, P(late) = 0.1. Find P(late).
Show answer
0.19. 0.3 × 0.4 + 0.7 × 0.1 = 0.12 + 0.07 = 0.19.
P(rain) = 0.3. If it rains, P(late) = 0.4; if not, P(late) = 0.1. Given that a person is late, find the probability it rained. Give your answer to 3 significant figures.
Show answer
0.632. P(rain | late) = 0.12/0.19 = 0.632.

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Last updated 7 November 2026. Spotted a mistake? Tell us.