Probability tree diagram questions

A probability tree diagram shows two or more events in order, with one set of branches for each event and the probability written on every branch. To find the probability of a path, multiply along the branches. To combine paths, add their results. These probability tree diagram questions start simple and build up to without replacement and algebra, with full answers.

How to use a tree diagram

In this tree, the probability of rain is 0.3. If it rains, the probability that Jo is late for school is 0.4; if it does not rain, it is 0.1.

1. Branches from one point add up to 1. 0.3 + 0.7 = 1, 0.4 + 0.6 = 1 and 0.1 + 0.9 = 1.
2. Multiply along a path for "this and then that". P(rain and late) = 0.3 × 0.4 = 0.12.
3. Add the paths you want for "either way". P(late) = P(rain and late) + P(no rain and late) = 0.12 + 0.07 = 0.19.
4. Check: the end results add to 1: 0.12 + 0.18 + 0.07 + 0.63 = 1.

AQA GCSE Maths covers tree diagrams for listing outcomes and for independent and dependent combined events at both tiers, including knowing when to add and when to multiply. Using a tree diagram for conditional probability is Higher tier.

Worked questions

Question 1: independent events

At a fair, Leo throws two hoops at a peg. Each hoop lands on the peg with probability 0.6, whatever happened to the other. Find the probability that (a) both land on, (b) exactly one lands on, (c) at least one lands on.

1. The first hoop does not affect the second, so the same probabilities go on both sets of branches: on (O) 0.6, miss (M) 0.4.
2. (a) P(O, O) = 0.6 × 0.6 = 0.36.
3. (b) Exactly one on can happen two ways: P(O, M) + P(M, O) = 0.6 × 0.4 + 0.4 × 0.6 = 0.48.
4. (c) At least one on is everything except two misses: 1 − 0.4 × 0.4 = 1 − 0.16 = 0.84.

Question 2: without replacement

A box has 10 chocolates: 3 dark and 7 milk. Ali eats two, one after the other, at random. Find the probability that (a) both are dark, (b) one of each kind.

1. First branch: P(dark) = 3/10, P(milk) = 7/10.
2. After a dark one is eaten, 9 are left and 2 are dark, so the second branches change: 2/9 dark, 7/9 milk. After a milk one: 3/9 dark, 6/9 milk.
3. (a) P(dark, dark) = 3/10 × 2/9 = 6/90 = 1/15.
4. (b) P(dark, milk) + P(milk, dark) = 3/10 × 7/9 + 7/10 × 3/9 = 21/90 + 21/90 = 42/90 = 7/15.

Without replacement, the second-stage fractions have a denominator one smaller, and the numerator drops for whichever colour was taken. Practise path calculations in this question from the Probability course.

Question 3: three events

A fair dice is rolled three times. Find the probability of getting at least one six.

1. Drawing every branch would give 8 paths, but only one path has no six at all: not six, not six, not six.
2. P(no sixes) = 5/6 × 5/6 × 5/6 = 125/216.
3. P(at least one six) = 1 − 125/216 = 91/216, about 0.421.

Question 4: using a tree to find a probability "given" (Higher)

Using the rain tree above: Jo is late. What is the probability that it rained?

• P(rain | late) = P(rain and late) ÷ P(late) = 0.12 ÷ 0.19 = 12/19, about 0.632.
• See conditional probability questions for more like this.

Question 5: tree diagrams with algebra (Higher)

A bag contains n sweets, of which 6 are orange. Two sweets are taken without replacement. The probability that both are orange is 1/3. Find n.

1. P(orange, orange) = 6/n × 5/(n − 1) = 30 ÷ (n(n − 1)).
2. Set it equal to 1/3: 30 ÷ (n(n − 1)) = 1/3, so n(n − 1) = 90.
3. n² − n − 90 = 0, which factorises to (n − 10)(n + 9) = 0.
4. n must be positive, so n = 10. Check: 6/10 × 5/9 = 30/90 = 1/3.

Tree diagram questions to try

1. A spinner lands on green with probability 0.7. It is spun twice. Find P(green both times).
2. For the same spinner, find P(at least one green) in two spins.
3. A bag holds 5 red and 7 blue marbles. Two are taken without replacement. Find P(both red) and P(one of each colour).
4. P(Sam catches the early bus) = 0.25. If he catches it, P(on time) = 0.8; if not, P(on time) = 0.6. Find P(Sam is on time).
5. Using the rain tree, find P(on time).

Answers

1. 0.7 × 0.7 = 0.49.
2. 1 − 0.3 × 0.3 = 1 − 0.09 = 0.91.
3. P(both red) = 5/12 × 4/11 = 20/132 = 5/33. P(one of each) = 5/12 × 7/11 + 7/12 × 5/11 = 70/132 = 35/66.
4. 0.25 × 0.8 + 0.75 × 0.6 = 0.2 + 0.45 = 0.65.
5. 0.18 + 0.63 = 0.81 (or 1 − 0.19).

Check your answers to questions like these in the Probability Calculator.

Common mistakes

• Adding along a path. Along a path you multiply; you only add separate paths.
• Forgetting a path. "One of each" usually means two paths, such as red then blue and blue then red.
• Not changing the fractions without replacement. Both the numerator and the denominator can change on the second branches.
• Branches that do not add to 1. Check every pair before you use the tree.
• Doing "at least one" the long way. 1 − P(none) is quicker and safer.

Practise with a short timed paper

Tree diagrams come up at both tiers of GCSE Maths, with conditional probability on Higher. The daily GCSE Maths Higher and GCSE Maths Foundation papers are short timed papers of 8 questions, with a new one every day.

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