Cambridge IGCSE Statistics: Probability distributions and expectation (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 Topic 8, Probability distributions (8.1 and 8.2) of Cambridge IGCSE Statistics (0479), from the Cambridge IGCSE Statistics 0479 syllabus for examination in 2027. Use the study guide first if you need the methods, or the revision notes for a quick refresher. The course hub, printable checklist and free diagnostics are also available. Show all working.

Questions

1. A spinner gives a score X. The distribution is shown below.

x	2	4	6	8
P(X = x)	0.35	0.30	0.25	0.10

(a) Show that this is a valid probability distribution. [1] (b) Calculate E(X). [2] (c) Find P(X > 4). [1]

2. A bag holds 2 gold and 4 silver beads. Two beads are taken at random without replacement. Let X be the number of gold beads taken.

(a) Find the probability distribution of X and present it in a table. [3] (b) Calculate the expectation of X. [2]

3. A fair spinner has three equal sectors numbered 1, 2 and 3. It is spun twice. Let X be the difference between the two scores, found by subtracting the smaller from the larger (so X is 0 if the scores are equal).

(a) Complete the probability distribution of X, using a table of outcomes. [3] (b) Calculate E(X). [2] (c) Find P(X ≥ 1). [1]

4. A factory makes bolts. Each bolt is faulty with probability 0.03, independently of the others. A batch contains 600 bolts.

(a) Calculate the expected number of faulty bolts in a batch. [2] (b) Each faulty bolt costs £4.50 to replace. Calculate the expected replacement cost for a batch. [1] (c) In one batch there are 23 faulty bolts. Does this mean the probability of 0.03 is wrong? Explain. [1]

5. At a fair you pay £3 to roll a fair six-sided die. If you roll a 6 you receive £10. If you roll a 4 or a 5 you receive £3. Otherwise you receive nothing. Let X be the amount you receive.

(a) Write down the probability distribution of X. [2] (b) Calculate E(X). [2] (c) Find the expected gain per game, and state whether the game is fair. Find the expected total gain after 120 games. [2]

6. A variable X takes the values 1, 2, 3 and 4. P(X = 1) = 0.2 and P(X = 4) = 0.1. Let a = P(X = 2) and b = P(X = 3). It is given that E(X) = 2.5.

(a) Find the values of a and b. [4] (b) Find P(X ≥ 3). [1] (c) X only takes whole-number values. Explain why E(X) = 2.5 is still possible. [1]

7. The random variable X takes the values 0, 1, 2 and 3, and P(X = x) = k(x² + 1).

(a) Show that k = 1/18. [3] (b) Calculate E(X). [2] (c) Find P(X < 2). [2]

8. An insurance company sells a policy for £40. In a year, the probability of a claim of £100 is 0.07 and the probability of a claim of £500 is 0.03. Otherwise there is no claim. Let X be the amount the company pays out on one policy.

(a) Find the probability that there is no claim. [1] (b) Calculate the expected payout on one policy. [2] (c) Calculate the company’s expected profit on one policy. [1] (d) The company sells 5000 policies. Calculate its expected total profit. [1] (e) What premium would make the expected profit on each policy zero? [1]

9. Ali and Bea each take one shot at a target. Ali hits with probability 0.7 and Bea hits with probability 0.5. Their shots are independent. Let X be the total number of hits.

(a) Find the probability distribution of X. [4] (b) Calculate E(X). [2] (c) The pair win £20 if both hit, £5 if exactly one hits and nothing otherwise. Calculate the expected prize. [2]

10. A bag contains 3 red, 2 blue and 1 green counter. You pay £4 to play. You take two counters at random without replacement. If both counters are the same colour you receive £9. If exactly one counter is green you receive £3. Otherwise you receive nothing.

(a) Show that 4/15 is the probability of taking two counters of one colour. [3] (b) Find the probability that exactly one counter is green. [2] (c) Complete the probability distribution of the amount received. [1] (d) Calculate the expected amount received. [2] (e) Find the expected gain per game. Find the expected total gain for a player who plays 150 times. [2]

Answers

1. (a) 0.35 + 0.30 + 0.25 + 0.10 = 1, and every probability is between 0 and 1, so the distribution is valid. [1] (b) E(X) = 2 × 0.35 + 4 × 0.30 + 6 × 0.25 + 8 × 0.10 = 0.7 + 1.2 + 1.5 + 0.8 [1] E(X) = 4.2 [1] (c) P(X > 4) = 0.25 + 0.10 = 0.35 [1] Examiner insight: For “show that”, write out the addition. Saying the total is 1 without it gives the reader nothing to check.

2. (a) P(X = 0) = 4/6 × 3/5 = 12/30 = 2/5 [1] P(X = 2) = 2/6 × 1/5 = 2/30 = 1/15 [1] P(X = 1) = 1 − 2/5 − 1/15 = 8/15 (or two routes: 2/6 × 4/5 + 4/6 × 2/5 = 16/30) [1]

Distribution: 6/15, 8/15, 1/15 for x = 0, 1, 2.

(b) E(X) = 1 × 8/15 + 2 × 1/15 = 10/15 [1] E(X) = 2/3 [1] Examiner insight: The two routes for X = 1 are easy to miss. If you find it by subtraction, check it against the two-route sum.

3. (a) The nine equally likely outcomes give these differences:

First \ Second	1	2	3
1	0	1	2
2	1	0	1
3	2	1	0

Counts: 3 zeros, 4 ones, 2 twos [1]

Distribution: 3/9, 4/9, 2/9 for x = 0, 1, 2.

Total 9/9 = 1 [1] Correct outcomes grid [1] (b) E(X) = (0 × 3 + 1 × 4 + 2 × 2)/9 = 8/9 [1] E(X) = 8/9 (about 0.889) [1] (c) P(X ≥ 1) = 4/9 + 2/9 = 2/3 [1] Examiner insight: A full grid of all nine outcomes lets you count rather than guess.

4. (a) Expected number = n × p = 600 × 0.03 [1] = 18 faulty bolts [1] (b) 18 × £4.50 = £81 [1] (c) No. Expectation is an average over many batches. One batch varies by chance, so 23 does not show 0.03 is wrong. [1] Examiner insight: A comment needs a reason linked to the context. A bare “no” is unlikely to earn credit.

5. (a) P(10) = 1/6, P(3) = 2/6, P(0) = 3/6 [1]

Table with total 1 [1] (b) E(X) = 10 × 1/6 + 3 × 2/6 = 10/6 + 6/6 [1] E(X) = 16/6 = 8/3, about £2.67 [1] (c) Expected gain = 8/3 − 3 = −1/3, so −£0.33 per game (to the nearest penny), and the game is not fair as the expected gain is not zero. [1] Over 120 games: 120 × (−1/3) = −£40 [1] Examiner insight: Subtract the stake: £2.67 answers the payout question, not the one about gain. Using the exact fraction 1/3 for 120 games avoids a rounding slip.

6. (a) a + b = 1 − 0.2 − 0.1 = 0.7 [1] E(X) = 1 × 0.2 + 2a + 3b + 4 × 0.1 = 2.5, so 2a + 3b = 1.9 [1] Substitute a = 0.7 − b: 1.4 − 2b + 3b = 1.9, so b = 0.5 [1] b = 0.5 and a = 0.2 [1] (b) P(X ≥ 3) = 0.5 + 0.1 = 0.6 [1] (c) E(X) is a weighted average, so it need not be a value X can take. [1] Examiner insight: Two unknowns need two equations, so state both. The check 0.2 + 0.4 + 1.5 + 0.4 = 2.5 catches slips.

7. (a) The values of x² + 1 are 1, 2, 5, 10 [1] k(1 + 2 + 5 + 10) = 18k = 1 [1] So k = 1/18 [1] (b) E(X) = (0 × 1 + 1 × 2 + 2 × 5 + 3 × 10)/18 = 42/18 [1] E(X) = 7/3 [1] (c) P(X < 2) = P(0) + P(1) = 1/18 + 2/18 [1] = 3/18 = 1/6 [1] Examiner insight: In a “show that” question the answer is given, so display every step. Note that P(X = 0) = k, not zero.

8. (a) 1 − 0.07 − 0.03 = 0.9 [1] (b) E(X) = 100 × 0.07 + 500 × 0.03 = 7 + 15 [1] = £22 [1] (c) Expected profit = 40 − 22 = £18 [1] (d) 5000 × 18 = £90 000 [1] (e) The premium would equal the expected payout: £22 [1] Examiner insight: The no-claim outcome adds nothing to E(X) but belongs in the distribution. Label clearly which quantity is payout and which is profit.

9. (a) P(X = 0) = 0.3 × 0.5 = 0.15 [1] P(X = 2) = 0.7 × 0.5 = 0.35 [1] P(X = 1) = 0.7 × 0.5 + 0.3 × 0.5 = 0.35 + 0.15 = 0.5 [1]

Distribution: 0.15, 0.50, 0.35 for x = 0, 1, 2.

Table, with total 1 shown [1] (b) E(X) = 1 × 0.5 + 2 × 0.35 = 0.5 + 0.7 [1] E(X) = 1.2 [1] (c) Expected prize = 20 × 0.35 + 5 × 0.5 = 7 + 2.5 [1] = £9.50 [1] Examiner insight: Multiplying along branches uses independence, so say so. Also 1.2 = 0.7 + 0.5, a quick check on part (b).

10. (a) P(RR) = 3/6 × 2/5 = 6/30 [1] P(BB) = 2/6 × 1/5 = 2/30 [1] P(same colour) = 8/30 = 4/15 [1] (b) P(green then not green) = 1/6 × 5/5 = 1/6, and P(not green then green) = 5/6 × 1/5 = 1/6 [1] Total = 1/3 [1] (c) P(nothing) = 1 − 4/15 − 5/15 = 6/15 = 2/5

Distribution: £9 with 4/15, £3 with 1/3, £0 with 2/5.

[1] (d) E = 9 × 4/15 + 3 × 1/3 = 2.4 + 1 [1] = £3.40 [1] (e) Expected gain = 3.40 − 4 = −£0.60 per game [1] Over 150 games: 150 × (−0.60) = −£90 [1] Examiner insight: “Same colour” and “exactly one green” cannot both occur, so the three rows are separate and add to 1. Check this before calculating E.

Where marks are usually lost

• Skipping the ΣP(x) = 1 check, so a wrong entry goes unnoticed.
• Missing one route when a value can occur in two ways, such as one red and one blue in either order.
• Writing E(X) as the simple average of the x values, ignoring the probabilities.
• Giving the expected payout when the question asks for the expected gain, or forgetting to subtract the stake.
• Forgetting to multiply by the number of plays for a total.
• Using n × p when the probability changes between trials, as it does without replacement.

Next steps

• Review the revision notes and the study guide.
• Visit the course hub and the printable checklist.
• Try the free 10-minute diagnostics to find your weakest topics.
• Book a free trial class if you want help with these topics.

Official syllabus

Cambridge IGCSE Statistics 0479 syllabus for examination in 2027, Cambridge University Press & Assessment (Cambridge International Education), Topic 8: Probability distributions, learning outcomes 8.1 and 8.2.