Pearson Edexcel International Advanced Level Mathematics: Discrete random variables (YMA01) – 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 are aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics (YMA01) specification, Issue 3, April 2019, unit S1 section 5 (topics 5.1 to 5.4). The specification states that S1 is an optional unit for IAS Mathematics and Further Mathematics and for IAL Mathematics and Further Mathematics.

Learn the methods first in the study guide or the revision notes. The course page is Edexcel A Level Mathematics and the printable list is the checklist. Marks in bold brackets are the marks for that part.

Questions

1. State, with a reason, whether each is a discrete random variable.

(a) The number of text messages a phone receives in one hour. [1] (b) The mass of a parcel in kilograms. [1] (c) The number shown when a spinner with equal sectors labelled 1 to 5 is spun. [1]

2. The probability function of X is p(x) = k(2x + 1) for x = 0, 1, 2, 3, 4, and zero otherwise.

(a) Find the value of k. [2] (b) Find P(X ≥ 3). [2]

3. The cumulative distribution function of X is shown.

x	1	2	3	4	5
F(x)	0.10	0.35	0.60	0.85	1

(a) Find p(3). [1] (b) Find P(X ≥ 3). [2] (c) Find the probability that X is even. [2]

4. The discrete random variable X has the distribution below, where a and b are constants.

x	1	2	3	4
p(x)	0.2	a	b	0.3

Given that E(X) = 2.5:

(a) find the values of a and b. [4] (b) find Var(X). [3]

5. E(X) = 2.2 and Var(X) = 1.6. The random variable Y is given by Y = 4 − 3X.

(a) Find E(Y). [2] (b) Find Var(Y). [2]

6. The discrete random variable X is uniform on the values 1, 2, …, n and Var(X) = 14.

(a) Find n. [3] (b) Write down E(X). [1] (c) Find P(X ≥ 10). [1]

7. The probability function of X is p(x) = kx² for x = 1, 2, 3, and zero otherwise.

(a) Show that k = 1/14. [2] (b) Find E(X). [2]

8. A fair six-sided die is rolled and the score is X.

(a) Show that E(X²) = 91/6. [2] (b) Hence find Var(X). [2] (c) A player pays £5 to play and receives £2X. Let T pounds be the player’s profit. Find E(T) and Var(T). [3]

9. The cumulative distribution function of X is F(x) = x(x + 1)/30 for x = 1, 2, 3, 4, 5.

(a) Show that p(x) = x/15 for x = 1, 2, 3, 4, 5. [3] (b) Find E(X). [2] (c) Find Var(X). [3]

10. A raffle ticket is drawn at random from tickets numbered 11, 12, …, 30. The number drawn is X.

(a) Write down P(X = x). [1] (b) Find E(X) and Var(X). [4] (c) Find P(X > 25) and P(15 ≤ X ≤ 24). [2]

11. A fair coin is tossed four times. Let X be the length of the longest run of identical faces, so HHTH has X = 2 and HHHH has X = 4.

(a) Show that p(1) = 1/8, p(2) = 1/2, p(3) = 1/4 and p(4) = 1/8. [3] (b) Find F(2) and hence P(X > 2). [2] (c) Find E(X). [2] (d) Find Var(X). [3]

12. The random variable X has E(X) = 6 and E(X²) = 40. The random variable Y = 5 − 2X.

(a) Find Var(X). [2] (b) Find E(Y). [1] (c) Find Var(Y). [1] (d) Show that E(Y²) = 65. [3]

Answers

1. (a) Discrete, because it takes whole-number values that can be listed 0, 1, 2, … [1] (b) Not discrete, because mass can take any value in an interval [1] (c) Discrete, because it takes only the values 1, 2, 3, 4, 5 [1] [3] Examiner insight: A bare “yes” or “no” is weak on a “state with a reason” question. A reason that mentions listable separate values, or any value in an interval, is what shows you understand the idea.

2. (a) Σ p(x) = 1: k(1 + 3 + 5 + 7 + 9) = 25k = 1 [1] k = 1/25 [1] (b) P(X ≥ 3) = p(3) + p(4) = 7/25 + 9/25 [1] = 16/25 [1] [4] Examiner insight: Write the equation Σ p(x) = 1 with every term visible. If you then slip in the arithmetic, the setup still shows your method.

3. (a) p(3) = F(3) − F(2) = 0.60 − 0.35 = 0.25 [1] (b) P(X ≥ 3) = 1 − F(2) [1] = 0.65 [1] (c) p(2) = 0.35 − 0.10 = 0.25 and p(4) = 0.85 − 0.60 = 0.25 [1] P(X is even) = 0.25 + 0.25 = 0.5 [1] [5] Examiner insight: “At least 3” includes 3, so the correct subtraction is F(2). Quoting 1 − F(3) is the commonest wrong turn, so say which values you are including.

4. (a) 0.2 + a + b + 0.3 = 1 gives a + b = 0.5 [1] E(X) = 0.2 + 2a + 3b + 1.2 = 2.5 gives 2a + 3b = 1.1 [1] Subtract 2(a + b) = 1.0: b = 0.1 [1] a = 0.4, b = 0.1 [1] (b) E(X²) = 0.2 + 4(0.4) + 9(0.1) + 16(0.3) = 0.2 + 1.6 + 0.9 + 4.8 = 7.5 [1] Var(X) = 7.5 − 2.5² [1] = 1.25 [1] [7] Examiner insight: Two unknowns need two equations: one from the probabilities adding to 1 and one from the mean. Check the answer by substituting back before you move to part (b).

5. (a) E(Y) = 4 − 3(2.2) [1] = −2.6 [1] (b) Var(Y) = (−3)² × 1.6 [1] = 14.4 [1] [4] Examiner insight: A negative multiplier is squared in the variance, so a negative variance signals an error. Writing the line (−3)² × 1.6 makes the sign handling visible.

6. (a) (n² − 1)/12 = 14 [1] n² = 169 [1] n = 13 (n is positive) [1] (b) E(X) = (13 + 1)/2 = 7 [1] (c) The values 10, 11, 12, 13 give P(X ≥ 10) = 4/13 [1] [5] Examiner insight: Reject the negative root of n² = 169 explicitly, since n counts values. Counting the values in the range, then dividing by n, is the safest way to a uniform probability.

7. (a) Σ p(x) = k(1 + 4 + 9) = 14k [1] 14k = 1, so k = 1/14 [1] (b) E(X) = (1/14)(1 × 1 + 2 × 4 + 3 × 9) [1] = 36/14 = 18/7 [1] [4] Examiner insight: On a “show that” the result is given, so every step to it must appear. Note that x² is in the probability, so E(X) multiplies each x by kx², giving cubes in the sum.

8. (a) E(X²) = (1² + 2² + 3² + 4² + 5² + 6²)/6 [1] = 91/6 [1] (b) E(X) = 7/2 [1] Var(X) = 91/6 − 49/4 = 35/12 [1] (c) T = 2X − 5, so E(T) = 2(7/2) − 5 = 2 [1] Var(T) = 2² × 35/12 [1] = 35/3 [1] [7] Examiner insight: The £5 fee shifts the mean and has no effect on the variance. Say so in a few words, because it shows the reason for leaving the 5 out.

9. (a) p(x) = F(x) − F(x − 1) = [x(x + 1) − (x − 1)x]/30 [1] = (x² + x − x² + x)/30 = 2x/30 [1] = x/15 for x = 2, 3, 4, 5, and p(1) = F(1) = 2/30 = 1/15 also fits [1] (b) E(X) = (1/15)(1 + 4 + 9 + 16 + 25) [1] = 55/15 = 11/3 [1] (c) E(X²) = (1/15)(1 + 8 + 27 + 64 + 125) = 225/15 = 15 [1] Var(X) = 15 − (11/3)² [1] = 15 − 121/9 = 14/9 [1] [8] Examiner insight: The first value of a cumulative table is a special case, since F(0) is not defined here. Checking that p(1) equals F(1) closes the argument.

10. (a) There are 20 equally likely values, so P(X = x) = 1/20 for x = 11, …, 30 [1] (b) Let Y = X − 10, uniform on 1 to 20 [1] E(X) = 21/2 + 10 = 20.5 [1] Var(Y) = (400 − 1)/12 = 399/12 [1] Var(X) = 33.25, unchanged by the shift [1] (c) P(X > 25) = 5/20 = 1/4 [1] P(15 ≤ X ≤ 24) = 10/20 = 1/2 [1] [7] Examiner insight: The values start at 11, so (n² − 1)/12 cannot be applied with n = 30. Showing the shift Y = X − 10 is a clear way to show your method here.

11. (a) The sample space has 16 equally likely outcomes [1] Longest run 1: only HTHT and THTH, so p(1) = 2/16 = 1/8. Longest run 4: HHHH and TTTT, so p(4) = 1/8 [1] Longest run 3: HHHT, THHH, TTTH, HTTT, so p(3) = 4/16 = 1/4. The remaining 8 outcomes have longest run 2, so p(2) = 1/2 [1] (b) F(2) = 1/8 + 1/2 = 5/8 [1] P(X > 2) = 1 − 5/8 = 3/8 [1] (c) E(X) = (1 × 2 + 2 × 8 + 3 × 4 + 4 × 2)/16 [1] = 38/16 = 19/8 [1] (d) E(X²) = (1 × 2 + 4 × 8 + 9 × 4 + 16 × 2)/16 = 102/16 = 51/8 [1] Var(X) = 51/8 − (19/8)² [1] = 408/64 − 361/64 = 47/64 [1] [10] Examiner insight: Listing the outcomes in groups, or counting them by a clear rule, is needed to justify each probability. Keep the fractions exact, since 47/64 is easier to check than a rounded decimal.

12. (a) Var(X) = 40 − 6² [1] = 4 [1] (b) E(Y) = 5 − 2(6) = −7 [1] (c) Var(Y) = (−2)² × 4 = 16 [1] (d) E(Y²) = Var(Y) + (E(Y))² [1] = 16 + (−7)² [1] = 65 [1] [7] Examiner insight: A second route, expanding E(25 − 20X + 4X²) = 25 − 120 + 160, is a good check. Using Var(Y) + (E(Y))² keeps the working short and shows the variance identity.

Where marks are usually lost

• Using 1 − F(a) instead of 1 − F(a − 1) when the question says “at least a”.
• Not writing the equation Σ p(x) = 1 before solving for a constant.
• Using E(X²) = (E(X))², or forgetting to square each x when finding E(X²).
• Subtracting E(X) rather than (E(X))² in the variance.
• Writing Var(aX + b) = a Var(X) + b.
• Using (n² − 1)/12 for a uniform variable that does not start at 1, without shifting.
• Taking n = 30 instead of 20 for values 11 to 30.
• Rounding E(X) early and using the rounded value in the variance.
• Leaving a “show that” with a missing line between the setup and the result.

Next steps

• Re-read the study guide for any topic you found hard.
• Use the revision notes as a last check.
• Open the Edexcel A Level Mathematics course page and tick off topics 5.1 to 5.4 on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class if you want help with these topics.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, first teaching September 2018, published by Pearson. These questions follow S1 section 5, topics 5.1 to 5.4.