IGCSE Mathematics 0580 · Topic 8.4

IGCSE Mathematics: Combined Probability and Tree Diagrams Practice Questions

Multiply along the branches of a tree diagram for events happening in sequence, and add between branches for alternative routes. Without replacement, both the numerator and the denominator change on the second draw.

Cambridge IGCSE Mathematics (0580) · Topic 8.4: Combined Probability and Tree Diagrams

Topic 8.4 of Cambridge IGCSE Mathematics 0580 is where the Extended paper separates candidates. The at least one question is almost always faster through the complement. The questions below cover replacement, non-replacement and that shortcut.

What you need to know for Combined Probability and Tree Diagrams

IGCSE Mathematics Combined Probability and Tree Diagrams questions and answers

4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[4 marks]

A bag contains 5 red and 7 blue counters. A counter is drawn, its colour noted, and replaced. A second counter is then drawn. Calculate the probability that both are red.

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Answer: 25 over 144
  1. There are 12 counters in total, so P(red) = 5 over 12 on the first draw.
  2. The counter is replaced, so the bag is unchanged and P(red) = 5 over 12 on the second draw too.
  3. The two draws are independent, so multiply along the branch.
  4. P(both red) = 5 over 12 x 5 over 12 = 25 over 144.
How the marks are awarded. 1 mark for P(red) = 5 over 12. 1 mark for the same probability on the second draw. 1 mark for multiplying. 1 mark for 25 over 144.
Where students lose the mark. Adding the two probabilities. Events happening one after the other are multiplied. Addition is for alternative outcomes.
Question 2[5 marks]

Using the same bag of 5 red and 7 blue counters, two counters are drawn without replacement. Calculate the probability that both are red.

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Answer: 5 over 33
  1. First draw: P(red) = 5 over 12.
  2. One red counter has been removed and not replaced, so 4 red counters remain out of a total of 11.
  3. Second draw: P(red) = 4 over 11.
  4. Multiply along the branch: 5 over 12 x 4 over 11 = 20 over 132.
  5. Simplify by dividing top and bottom by 4: 20 over 132 = 5 over 33.
How the marks are awarded. 1 mark for P = 5 over 12 on the first draw. 1 mark for reducing the reds to 4. 1 mark for reducing the total to 11. 1 mark for multiplying. 1 mark for 5 over 33.
Where students lose the mark. Changing the denominator to 11 but leaving the numerator at 5. Both numbers must change, because the counter removed was red.
Question 3[4 marks]

Using the same bag of 5 red and 7 blue counters, two counters are drawn without replacement. Calculate the probability that at least one is red.

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Answer: 15 over 22
  1. At least one red means one red or two reds, so working directly would need three separate paths.
  2. It is faster to use the complement: at least one red is the opposite of no reds at all.
  3. P(no reds) means both are blue: 7 over 12 x 6 over 11 = 42 over 132.
  4. P(at least one red) = 1 minus 42 over 132 = 90 over 132, which simplifies to 15 over 22.
How the marks are awarded. 1 mark for using the complement. 1 mark for P(both blue) = 7 over 12 x 6 over 11. 1 mark for 42 over 132. 1 mark for 15 over 22.
Where students lose the mark. Calculating only the probability of exactly one red. At least one includes the case where both are red, which is why the complement method is safer.
Question 4[3 marks]

The probability that it rains on a given day is 0.6, and independently the probability that a particular bus is late is 0.3. Calculate the probability that it rains and the bus is late.

Show the worked answer
Answer: 0.18
  1. The two events are stated to be independent, so one does not affect the other.
  2. For independent events, P(A and B) = P(A) x P(B).
  3. P(rain and late) = 0.6 x 0.3.
  4. P(rain and late) = 0.18.
How the marks are awarded. 1 mark for identifying that the events are independent and must be multiplied. 1 mark for correct substitution. 1 mark for 0.18.
Where students lose the mark. Adding to get 0.9. Addition applies to alternative outcomes, not to two things both happening.

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Combined Probability and Tree Diagrams FAQs

When do I multiply and when do I add probabilities?

Multiply along the branches of a tree when events happen one after another, meaning both occur. Add between branches when there are alternative ways of satisfying the condition, meaning one outcome or another.

What changes when there is no replacement?

Both the total number of items and the count of the type removed decrease by one for the second draw. Taking a red counter from a bag of 5 red and 7 blue leaves 4 red out of 11, so the second probability is 4 over 11, not 5 over 11.

How do I answer an at least one question?

Use the complement. Calculate the probability that none of the events happen, then subtract from 1. Working out every individual case that satisfies at least one is slower and risks missing a path.

What are independent events?

Events where the outcome of one has no effect on the other, such as rainfall and a bus running late. For independent events the probability of both happening is the product of the two individual probabilities.

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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.