Cambridge O Level Statistics: Discrete probability distributions, expectation and fair games (4040) – 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 12 of the Cambridge O Level Statistics 4040 syllabus for examination in 2025, 2026 and 2027: section 12.1 (probability distribution of a discrete variable) and section 12.2 (expectation, expected profit and loss, fair games). The syllabus is not tiered. Learn the methods first from the study guide and the revision notes. The course page is O Level Statistics and the printable list is the syllabus checklist. Marks in the answers are shown as [1] after each step.

Questions

1. A discrete variable X has the distribution below. Find k, then find P(X ≥ 2).

x	0	1	2	3
p(x)	0.10	0.30	k	0.35

[3]

2. The variable X takes the values 0, 1, 2, 3 with p(x) = k(x + 1).

(a) Find k. [2] (b) Find E(X). [2]

3. A drawer holds 4 black socks and 6 white socks. Two are taken at random without replacement. X is the number of black socks taken.

(a) Form the probability distribution of X. [3] (b) Calculate E(X). [2]

4. A survey of 60 households records the number of cars each owns.

Cars	0	1	2	3
Households	9	27	18	6

(a) Use the data to estimate the probability distribution of the number of cars, X. [2] (b) Calculate E(X). [2] (c) Explain what E(X) tells you, given that no household owns 1.35 cars. [1]

5. A variable X takes the values 1, 2, 3, 4 with p(1) = 0.15, p(2) = a, p(3) = b and p(4) = 0.25. Given that E(X) = 2.7, find a and b. [4]

6. The variable X takes the values 1, 2, 3 with probabilities 0.4, 0.35 and 0.25.

(a) Find E(X). [2] (b) Find E(X²). [2] (c) Find E(3X − 1). [2]

7. A game costs Rs 3 to play. A fair spinner is numbered 1 to 5. A score of 5 pays Rs 10, a score of 4 pays Rs 4, and any other score pays nothing.

(a) Write down the distribution of the net gain. [2] (b) Find the expected net gain per play. [1] (c) Find the expected total net gain over 150 plays. [1] (d) Is the game fair? Give a reason. [1]

8. A raffle sells 500 tickets. There is one prize of Rs 200, four prizes of Rs 50 and ten prizes of Rs 10.

(a) Find the ticket price that makes the raffle fair. [3] (b) The organiser sells every ticket at Rs 1.50. Find the organiser’s expected profit. [2]

9. To play a game you pay Rs 4 and take one card at random from 4 cards. One card wins Rs w, and the other three win nothing. Find w for the game to be fair. [4]

10. A fair coin is tossed up to three times. Tossing stops as soon as a head appears, or after the third toss. X is the number of tosses made.

(a) Find the distribution of X. [2] (b) Find E(X). [2] (c) A player pays Rs 2 and receives Rs 1 if the game stops on toss 1, Rs 3 if it stops on toss 2, and Rs 8 if it stops on toss 3. Find the expected net gain. [2] (d) The organiser changes only the Rs 8 prize so that the game is fair. Find the new prize. [2]

11. Two fair dice are rolled. Y is the positive difference between the two scores (0 if they match).

(a) Show that the distribution of Y is as follows. [3]

y	0	1	2	3	4	5
p(y)	6/36	10/36	8/36	6/36	4/36	2/36

(b) Calculate E(Y). [2] (c) A player pays Rs 2 to roll and receives Rs y. Find the expected net gain per play. [1] (d) Find the expected total net gain over 360 plays and say who benefits. [2]

Answers

1. Probabilities sum to 1: k = 1 − (0.10 + 0.30 + 0.35) [1] = 0.25 [1]. P(X ≥ 2) = 0.25 + 0.35 = 0.6 [1] [3] Examiner insight: Show the sum equal to 1 before the value of k; a bare k with no equation gives no way to see where a slip happened.

2. (a) The probabilities are k, 2k, 3k, 4k, so 10k = 1 [1], giving k = 1/10 [1]. (b) E(X) = 0 + 1 × 0.2 + 2 × 0.3 + 3 × 0.4 [1] = 0.2 + 0.6 + 1.2 = 2 [1] [4] Examiner insight: Writing each product x p(x) separately shows the method, so one slip does not hide the rest of your working.

3. (a) p(0) = 6/10 × 5/9 = 1/3 [1]. p(1) = 2 × (4/10 × 6/9) = 8/15 [1]. p(2) = 4/10 × 3/9 = 2/15 [1]. (b) E(X) = 8/15 + 2 × 2/15 [1] = 12/15 = 4/5 [1] [5] Examiner insight: A value reachable in two orders needs both branches added; a total of 1 in your table is the quickest check that you did this.

4. (a) Divide by 60: p(0) = 0.15, p(1) = 0.45 [1], p(2) = 0.30, p(3) = 0.10 [1]. (b) E(X) = 0.45 + 0.60 + 0.30 [1] = 1.35 [1] (c) In the long run the average number of cars per household is 1.35; it need not be a possible value [1] [5] Examiner insight: An interpretation needs the context (cars per household) and the idea of an average, not a statement that a household owns 1.35 cars.

5. Total: 0.15 + a + b + 0.25 = 1, so a + b = 0.6 [1]. Expectation: 0.15 + 2a + 3b + 1.0 = 2.7, so 2a + 3b = 1.55 [1]. Substitute a = 0.6 − b: 1.2 + b = 1.55 [1]. b = 0.35, a = 0.25 [1] [4] Examiner insight: Use both conditions: the total of 1 and the given expectation; using only one leaves two unknowns and cannot succeed.

6. (a) E(X) = 0.4 + 0.70 + 0.75 [1] = 1.85 [1] (b) E(X²) = 1 × 0.4 + 4 × 0.35 + 9 × 0.25 [1] = 4.05 [1] (c) E(3X − 1) = 3 × 1.85 − 1 [1] = 4.55 [1] [6] Examiner insight: Squaring 1.85 gives 3.4225, not 4.05; state clearly that you are squaring each value first.

7. (a) Net gain 7 with probability 1/5, net gain 1 with probability 1/5, net gain −3 with probability 3/5 [1] [1] (b) E = (7 + 1 − 9)/5 = −0.2, a loss of Rs 0.20 per play [1] (c) 150 × (−0.2) = −Rs 30, an expected loss of Rs 30 [1] (d) No, the expected net gain is not zero [1] [5] Examiner insight: Give the sign and the unit, and say whether it is a gain or a loss for the player; a bare negative number is easy to misread.

8. (a) Total prizes = 200 + 4 × 50 + 10 × 10 [1] = 500 [1]. Expected prize per ticket = 500/500, so the fair price is Rs 1 [1]. (b) Expected profit per ticket = 1.50 − 1 = Rs 0.50 [1]. For 500 tickets: Rs 250 [1] [5] Examiner insight: Explain that a fair price equals the expected prize per ticket; the figure alone does not show you understand why.

9. Net gain is w − 4 with probability 1/4 and −4 with probability 3/4 [1]. Expected net gain = (w − 4)/4 − 12/4 [1]. Fair, so (w − 4)/4 − 3 = 0 [1]. w = 16 [1] [4] Examiner insight: Set the expectation equal to zero explicitly; the answer 16 then follows by checking that 16 × 1/4 equals the stake.

10. (a) p(1) = 1/2, p(2) = 1/4 [1], p(3) = 1/4 (two tails, then stop whatever the third toss shows) [1]. (b) E(X) = 1/2 + 2/4 + 3/4 [1] = 7/4 = 1.75 [1] (c) E(received) = 1 × 1/2 + 3 × 1/4 + 8 × 1/4 = 3.25 [1]; net gain = 3.25 − 2 = Rs 1.25 in the player’s favour [1] (d) Let the prize be w: 1/2 + 3/4 + w/4 = 2 [1], so w = 3, a prize of Rs 3 [1] [8] Examiner insight: The stopping rule means the probability of 3 tosses is 1/4, not 1/8; check the three probabilities add to 1 before going on.

11. (a) There are 36 equally likely outcomes [1]. Count those with each difference: 0 occurs 6 ways, 1 occurs 10 ways, 2 occurs 8 ways, 3 occurs 6 ways, 4 occurs 4 ways, 5 occurs 2 ways [1]. The counts total 36, so each probability is the count over 36 [1]. (b) E(Y) = (0 + 10 + 16 + 18 + 16 + 10)/36 [1] = 70/36 = 35/18 ≈ 1.94 [1] (c) Expected net gain = 35/18 − 2 = −1/18 ≈ −Rs 0.06 per play [1] (d) 360 × (−1/18) = −Rs 20, an expected loss of Rs 20 for the player, so the organiser benefits [1] [1] [8] Examiner insight: A “show that” needs the counting displayed, for example a grid of differences; stating the final table alone gains little.

Where marks are usually lost

• Not checking that the probabilities total 1.
• Missing the second order when a value can occur in two ways.
• Using the with-replacement fractions when the question says without replacement.
• Dividing Σ x p(x) by the number of values.
• Writing E(X²) as the square of E(X).
• Treating the prize as the net gain instead of subtracting the stake.
• Writing an expected loss with no sign, no unit or no statement of who loses.
• Calling a game fair or unfair with no calculation of the expectation.
• Reading an expectation as the result of the next single play.

Next steps

• Re-read the revision notes and the study guide for any method you missed.
• Check your coverage on the O Level Statistics course page and the syllabus checklist.
• Take one of the free 10-minute diagnostics to find weak topics.
• Book a free trial class if you want help with any of these topics.

Official syllabus

Cambridge O Level Statistics 4040, syllabus for examination in 2025, 2026 and 2027, Cambridge University Press & Assessment (Cambridge International Education). Topic 12, sections 12.1 and 12.2, and the mathematical notation list.