A Level Further Maths chapter 4: Further Probability & Statistics MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

143 exam-style questions on Further Probability & Statistics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

143 questions · 5 topics · 10 per test

Test this chapter freeAll A Level Further Maths chapters

What chapter 4 covers

Ten Further Probability & Statistics questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. X has probability generating function G(t). Which expression gives Var(X)?

  1. A
    G″(1) − [G′(1)]2
  2. B
    G″(1) + [G′(1)]2
  3. C
    G″(1) − G′(1)
  4. D
    G″(1) + G′(1) − [G′(1)]2
Show answer

Answer: D. G″(1) = E(X(X − 1)) = E(X2) − E(X), so E(X2) = G″(1) + G′(1). Then Var(X) = E(X2) − [E(X)]2 = G″(1) + G′(1) − [G′(1)]2.

2. X has probability generating function G(t) = e3(t − 1). Find Var(X).

  1. A
    3
  2. B
    9
  3. C
    12
  4. D
    6
Show answer

Answer: A. G′(t) = 3e3(t − 1), so G′(1) = 3, and G″(t) = 9e3(t − 1), so G″(1) = 9. Var(X) = 9 + 3 − 9 = 3. The value 9 is G″(1) and 12 is G″(1) + G′(1), each missing the last step.

3. The probability generating function of X is G(t) = 0.2 + 0.5t + 0.3t3. Find Var(X).

  1. A
    3.2
  2. B
    0.4
  3. C
    1.8
  4. D
    1.24
Show answer

Answer: D. G′(t) = 0.5 + 0.9t2, so G′(1) = 1.4. G″(t) = 1.8t, so G″(1) = 1.8. Var(X) = G″(1) + G′(1) − [G′(1)]2 = 1.8 + 1.4 − 1.96 = 1.24. Omitting the last term gives 3.2.

4. The random variable X has probability density function f(x) = e−x for x ≥ 0, and f(x) = 0 otherwise. Find E(eX/2).

  1. A
    2/3
  2. B
    1/2
  3. C
    1
  4. D
    2
Show answer

Answer: D. E(eX/2) = ∫0∞ ex/2 e−x dx = ∫0∞ e−x/2 dx = [−2e−x/2]0∞ = 2. The exponents combine by subtraction, giving −x/2, not −3x/2 (which would give 2/3).

5. A geneticist crosses plants and expects offspring in the phenotype ratio 9 : 3 : 3 : 1. From 160 offspring, what is the expected frequency of the least common phenotype?

  1. A
    10
  2. B
    40
  3. C
    1
  4. D
    16
Show answer

Answer: A. The ratio parts sum to 9 + 3 + 3 + 1 = 16, so the least common phenotype has probability 1/16 and expected frequency 160/16 = 10.

6. The result GX+Y(t) = GX(t) GY(t) holds provided that:

  1. A
    X and Y have the same distribution
  2. B
    X and Y have the same mean
  3. C
    X and Y are both greater than zero
  4. D
    X and Y are independent
Show answer

Answer: D. The product rule follows from E(tX+Y) = E(tXtY) = E(tX)E(tY), and the last step requires independence. It does not need identical distributions.

7. The random variable X has probability density function f(x) = x/2 for 0 ≤ x ≤ 2, and f(x) = 0 otherwise. Find Var(3X − 1).

  1. A
    2
  2. B
    2/3
  3. C
    2/9
  4. D
    1
Show answer

Answer: A. E(X) = ∫ x · x/2 dx = 4/3 and E(X2) = ∫ x2 · x/2 dx = 2, so Var(X) = 2 − 16/9 = 2/9. Then Var(3X − 1) = 32 Var(X) = 2. The constant −1 has no effect, and the multiplier is squared.

8. A Poisson distribution is fitted to data with the mean estimated from the sample. After combining classes there are 5 classes, and the test statistic is χ2 = 8.4. The table values for the 5% significance level are 7.815 (ν = 3) and 9.488 (ν = 4). What is the conclusion?

  1. A
    Do not reject H0, since 8.4 < 9.488
  2. B
    Do not reject H0, since 8.4 < 11.07
  3. C
    Do not reject H0, since 8.4 > 7.815
  4. D
    Reject H0, since 8.4 > 7.815
Show answer

Answer: D. Here ν = 5 − 1 − 1 = 3, because the mean was estimated. The critical value is 7.815 and 8.4 exceeds it, so H0 (Poisson fits) is rejected. Using ν = 4 (9.488) would wrongly give the opposite conclusion.

9. 200 observations are compared with a Poisson distribution with mean 2. The classes are 0, 1, 2, 3, 4 and "5 or more". What is the expected frequency of the "5 or more" class?

  1. A
    10.5
  2. B
    7.22
  3. C
    28.6
  4. D
    3.31
Show answer

Answer: A. P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.9473 = 0.0527, so the expected frequency is 200 × 0.0527 = 10.5. The value 7.22 is P(X = 5) only, and the others use the wrong cut-off.

10. The random variable X takes the values 1, 2, 3, 4 and 5 with equal probabilities. Which is its probability generating function?

  1. A
    (1 + t + t2 + t3 + t4)/5
  2. B
    (t + t2 + t3 + t4 + t5)/5
  3. C
    t + t2 + t3 + t4 + t5
  4. D
    t5/5
Show answer

Answer: B. Each value x has probability 1/5, so G(t) = Σ (1/5) tx = (t + t2 + t3 + t4 + t5)/5. Starting from t0 would represent the values 0 to 4, and G(1) must equal 1.

Start the chapter testAbout the A Level Further Maths test

Read the notes for this chapter

Ways to practise this chapter

Further Probability & Statistics: questions students ask

How many Further Probability & Statistics MCQs are there?

143, each with the answer and a short explanation, across 5 topics: Continuous random variables, Inference using normal and t-distributions, χ²-tests, Non-parametric tests, Probability generating functions. Every chapter test draws ten of them, so you can take it many times.

Is the A Level Further Maths chapter 4 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Keep going

← Chapter 3: Further MechanicsAll chaptersStuck? Ask a question