Cambridge IGCSE Statistics: Probability basics: notation, simple cases with and without replacement, complements and the addition rule (0479) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover outcomes 7.1 to 7.4 of Topic 7 (Probability) in the Cambridge IGCSE Statistics (0479) syllabus for examination in 2027. If a method is unfamiliar, read the study guide or the revision notes first. The course hub and printable checklist show the full course. Show your working throughout. Give fractions in their simplest form unless a question asks for another form.

Questions

1. For a randomly chosen pupil, A is the event “walks to school” and B is the event “has a packed lunch”. Write each of the following using probability notation.

(a) The pupil does not walk to school. [1] (b) The pupil walks to school and has a packed lunch. [1] (c) The pupil walks to school or has a packed lunch (or both). [1]

2. The ten letters of the word PERCENTILE are written on identical tiles and placed in a bag. One tile is chosen at random.

(a) Find the probability that it is an E. Give your answer as a fraction. [1] (b) Find the probability that it is a vowel. Give your answer as a decimal. [1] (c) Find the probability that it is not an E. Give your answer as a percentage. [1]

3. Find the probability of the complement in each case.

(a) The probability that it rains on a given day is 0.35. Find the probability that it does not rain. [1] (b) The probability that a match has no goals is 5/12. Find the probability that it has at least one goal. [1] (c) P(A) = 64%. Find P(A′) as a percentage. [1]

4. Two fair dice are rolled and the scores are added.

(a) Find the probability that the total is 8. [2] (b) Find the probability that the total is more than 3. [2]

5. A bag holds 7 yellow and 3 green counters. Two counters are taken one after the other.

(a) The first counter is replaced before the second is taken. Find the probability that both are green. [2] (b) The first counter is not replaced. Find the probability that both are green. [2] (c) State why the answer to (b) differs from the answer to (a). [1]

6. A hat holds the names of 5 girls and 4 boys. Two names are drawn without replacement.

(a) Find the probability that one name is a girl’s and one is a boy’s. [3] (b) Find the probability that both names are of the same gender. [2]

7. A shelf holds 9 mugs and 4 of them are chipped. Three mugs are taken without replacement.

(a) Find the probability that none of the three is chipped. [2] (b) Find the probability that at least one is chipped. [2]

8. For two events, P(A) = 0.48, P(B) = 0.35 and P(A ∩ B) = 0.12.

(a) Find P(A ∪ B). [2] (b) Find the probability that neither A nor B happens. [1] (c) Find P(A ∩ B′). [1]

9. For two events, P(A) = 0.64, P(B) = 0.38 and P(A ∪ B) = 0.85.

(a) Find P(A ∩ B). [2] (b) Find the probability that B happens but A does not. [1] (c) Find P(A′ ∩ B′). [1]

10. A library has 120 registered readers. Of these, 70 have borrowed fiction (F), 55 have borrowed non-fiction (N) and 25 have borrowed both. A reader is chosen at random.

(a) Find P(F). [1] (b) Find P(F ∩ N). [1] (c) Find P(F ∪ N). [2] (d) Find the probability that the reader has borrowed neither. [1] (e) Two different readers are chosen at random. Find the probability that neither has borrowed either. [2]

11. A bag holds 6 red, 4 blue and 2 green counters. Two counters are taken without replacement.

(a) Find the probability that both are red. [2] (b) Find the probability that the two counters are different colours. [4] (c) Find the probability that at least one counter is green. [2]

Answers

1. (a) P(A′) [1] (b) P(A ∩ B) [1] (c) P(A ∪ B) [1] [3] Examiner insight: Learn the symbols as pairs: ∩ goes with “and” and ∪ goes with “or”, and writing the word beside the symbol in your working helps you avoid swapping them.

2. (a) E appears 3 times in 10 letters, so P(E) = 3/10 [1] (b) The vowels are E, E, I and E, which is 4 tiles, so P(vowel) = 4/10 = 0.4 [1] (c) There are 3 E tiles, so P(not E) = 1 – 3/10 = 7/10 = 70% [1] [3] Examiner insight: Count repeated letters one tile at a time. A common slip is to count the letter types (8 different letters) instead of the 10 tiles.

3. (a) 1 – 0.35 = 0.65 [1] (b) 1 – 5/12 = 7/12 [1] (c) 100% – 64% = 36% [1] [3] Examiner insight: Keep the form the question uses. Switching between decimal, fraction and percentage in the middle of an answer invites arithmetic slips and can leave the final form unclear.

4. (a) There are 36 equally likely pairs. The pairs with total 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), so 5 pairs. [1] P = 5/36 [1] (b) A total of 3 or less comes from (1,1), (1,2) and (2,1), which is 3 pairs. [1] P(more than 3) = 1 – 3/36 = 33/36 = 11/12 [1] [4] Examiner insight: A written list or table of the outcomes shows your counting. A bare fraction with a wrong count gives the reader nothing to credit.

5. (a) Each pick is from 10 counters, so P(both green) = 3/10 × 3/10 [1] = 9/100 = 0.09 [1] (b) P(both green) = 3/10 × 2/9 [1] = 6/90 = 1/15 [1] (c) After a green counter is kept out, only 2 green counters remain among 9 counters, so the second pick has a smaller chance of green. [1] [5] Examiner insight: For a “state why” part, link the change in the numbers to the context. “The numbers change” alone is vague, while “2 green left out of 9” is specific.

6. (a) Girl then boy: 5/9 × 4/8 = 20/72. [1] Boy then girl: 4/9 × 5/8 = 20/72. [1] P(one of each) = 40/72 = 5/9 [1] (b) “Same gender” is the complement of “one of each”. [1] P = 1 – 5/9 = 4/9 [1] [5] Examiner insight: If the question gives no order, include both orders. Check part (b) by adding girl-girl 20/72 and boy-boy 12/72, which also gives 32/72 = 4/9.

7. (a) 5 mugs are not chipped. P(none chipped) = 5/9 × 4/8 × 3/7 [1] = 60/504 = 5/42 [1] (b) “At least one” is the complement of “none”. [1] P = 1 – 5/42 = 37/42 [1] [4] Examiner insight: Reduce the numerator and denominator by one at each pick. Keeping 9 as the total on every pick is a frequent error without replacement.

8. (a) P(A ∪ B) = 0.48 + 0.35 – 0.12 [1] = 0.71 [1] (b) P(neither) = 1 – 0.71 = 0.29 [1] (c) P(A ∩ B′) = 0.48 – 0.12 = 0.36 [1] [4] Examiner insight: Write the rule first and then substitute. Showing the subtracted overlap makes it clear you did not simply add the two probabilities.

9. (a) Rearranging the rule, P(A ∩ B) = P(A) + P(B) – P(A ∪ B) [1] = 0.64 + 0.38 – 0.85 = 0.17 [1] (b) P(B ∩ A′) = 0.38 – 0.17 = 0.21 [1] (c) P(A′ ∩ B′) = 1 – 0.85 = 0.15 [1] [4] Examiner insight: When a union is given, rearrange the rule before you substitute. Check that your overlap is no larger than either single probability.

10. (a) P(F) = 70/120 = 7/12 [1] (b) P(F ∩ N) = 25/120 = 5/24 [1] (c) P(F ∪ N) = 70/120 + 55/120 – 25/120 [1] = 100/120 = 5/6 [1] (d) P(neither) = 1 – 5/6 = 1/6 [1] (e) 20 readers have borrowed neither, so P(first has borrowed neither) = 20/120 and then P(second has borrowed neither) = 19/119. [1] P = 20/120 × 19/119 = 19/714 [1] [7] Examiner insight: Use the earlier parts. Part (d) gives 20 readers from 120, and for (e) the numerator and denominator both drop by one, because the first reader is not put back.

11. (a) P(both red) = 6/12 × 5/11 [1] = 30/132 = 5/22 [1] (b) Same colour: blue 4/12 × 3/11 = 12/132 and green 2/12 × 1/11 = 2/132. [1] With red 30/132, P(same) = 44/132 = 1/3. [1] Different colours is the complement. [1] P = 1 – 1/3 = 2/3 [1] (c) P(no green) = 10/12 × 9/11 = 90/132. [1] P(at least one green) = 1 – 90/132 = 42/132 = 7/22 [1] [8] Examiner insight: For “different colours” with three colours, finding “same colour” and subtracting from 1 needs three short products instead of six. Write each product clearly so the sum is easy to follow.

Where marks are usually lost

• Treating “A or B” as a sum of the two separate probabilities, with no allowance for the overlap.
• Mixing up ∩ and ∪ in notation questions.
• Using the full total on every pick when the question says “without replacement”.
• Counting only one order for “one of each” or “different colours”.
• Counting letter types instead of tiles when letters repeat, as in question 2.
• Using a long list for “at least one” when the complement is quicker.
• Not simplifying a fraction, or giving a decimal or percentage when the question asked for another form.
• Writing only a final answer. The syllabus says essential working must be shown for full marks to be awarded.

Next steps

• Revise with the revision notes and re-read the study guide.
• Open the Cambridge IGCSE Statistics hub and tick off the outcomes on the checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Cambridge IGCSE Statistics 0479 syllabus for examination in 2027, Version 1, Cambridge International Education (part of Cambridge University Press & Assessment). These questions cover Topic 7, outcomes 7.1 to 7.4.