IGCSE Maths probability: tree diagrams, Venn diagrams and conditional probability

What each tier expects

The key ideas

For equally likely outcomes, P(event) = number of favourable outcomes / total number of outcomes. A sample space diagram lists every outcome, which is the safest way to handle two dice or two spinners: with two fair dice there are 36 outcomes and 6 give a total of 7, so P(total 7) = 6/36 = 1/6.

Relative frequency, the number of times an event happened divided by the number of trials, estimates a probability from an experiment, and it becomes more reliable as the number of trials grows. Expected frequency works the other way: if P(event) = 0.4 and there are 300 trials, expect 0.4 × 300 = 120 occurrences.

On a tree diagram, each set of branches from a point adds to 1. To find the probability of a route, multiply along it. To find the probability of an event made of several routes, add those routes. If items are not replaced, the second-stage fractions change: both the number of that colour and the total go down by one.

Conditional probability (Cambridge Extended, Edexcel Higher) is the probability of A given that B has happened. Restrict your attention to B: from a Venn diagram, it is the number in both A and B divided by the number in B.

Worked example 1: two counters without replacement

1. A bag has 4 red and 3 blue counters. Two are taken at random without replacement. Find P(both red) and P(one of each colour).
2. First pick: P(R) = 4/7, P(B) = 3/7. If the first is red, 6 counters remain with 3 red, so P(R second) = 3/6. If the first is blue, P(R second) = 4/6.
3. P(both red) = 4/7 × 3/6 = 12/42 = 2/7.
4. P(one of each) = P(R then B) + P(B then R) = 4/7 × 3/6 + 3/7 × 4/6 = 12/42 + 12/42 = 24/42 = 4/7.
5. Check: P(both blue) = 3/7 × 2/6 = 6/42 = 1/7, and 2/7 + 4/7 + 1/7 = 1. All outcomes accounted for.

Worked example 2: Venn diagram and conditional probability

1. In a group of 40 students, 25 like tea, 18 like coffee and 8 like both.
2. Fill the Venn diagram from the middle: both = 8, tea only = 25 - 8 = 17, coffee only = 18 - 8 = 10, neither = 40 - 17 - 8 - 10 = 5.
3. P(likes tea but not coffee) = 17/40.
4. Given a student likes coffee, the probability they also like tea: restrict to the 18 coffee drinkers, of whom 8 like tea. P = 8/18 = 4/9.

Worked example 3: expected frequency and relative frequency

1. A spinner landed on red 32 times in 80 spins. Estimate P(red).
2. Relative frequency = 32/80 = 0.4.
3. How many reds would you expect in 200 spins? 0.4 × 200 = 80.
4. Is the spinner fair if it has four equal sections, one red? A fair spinner gives P(red) = 0.25. An estimate of 0.4 from 80 spins suggests bias, but more trials would give a more reliable conclusion.

Common mistakes that cost marks

• Adding along branches instead of multiplying.
• Keeping the same denominator on the second branch when there is no replacement.
• Finding only one route for 'one of each' when there are two orders.
• Giving a probability as a ratio (4 : 7) or in words ('4 out of 7') instead of a fraction, decimal or percentage.
• Double-counting the overlap in a Venn diagram by putting 25 in the tea-only region.
• Probabilities greater than 1 or negative, which should always trigger a recheck.

Exam technique and how a tutor helps

Draw the tree diagram even when it is not asked for, and write the probabilities on the branches and the outcomes at the ends, which is the layout Cambridge uses. Leave fractions unsimplified until the end; it makes adding routes easier because the denominators already match. On calculator papers a fraction answer is still usually expected unless the question says otherwise.

For 'at least one' questions, use the complement: P(at least one red) = 1 - P(no reds). It is quicker and less error-prone than adding several routes.

Probability questions often hinge on reading the context correctly, such as whether items are replaced, or whether the question means 'given that'. In one-to-one lessons a tutor has the student underline those words, sketch the tree together on the shared whiteboard, and then explain each branch aloud. That habit catches most of the errors that cost marks here.

Self-check: can you do these?

• P(B) = 0.8. Find P(not B). (Answer: 0.2)
• Two fair coins are thrown. Find P(at least one head). (Answer: 3/4)
• A bag has 6 green and 4 yellow balls. Two are taken without replacement. Find P(both yellow). (Answer: 2/15)
• The same bag, with replacement. Find P(both green). (Answer: 0.36)
• An event has probability 0.15. How many times would you expect it in 400 trials? (Answer: 60)