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IB · MATH AA HL

Mathematics AA: Higher Level

Statistics and probability — Topic 4

Name: ____________________Date: October 10, 2026
  1. 1.

    X ~ B(10, 0.3). Find P(X = 3).

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States the binomial probability formula P(X = k) = ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes n = 10, k = 3, p = 0.3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates ¹⁰C₃ = 120. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates (0.3)³(0.7)⁷ ≈ 0.002224. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains P(X = 3) ≈ 0.267. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: On the exam this is usually read directly from calculator binomial-probability functions, but the underlying formula must still be stated for method marks.

    Marking points

    • States the binomial probability formula P(X = k) = ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ.
    • Substitutes n = 10, k = 3, p = 0.3.
    • Calculates ¹⁰C₃ = 120.
    • Calculates (0.3)³(0.7)⁷ ≈ 0.002224.
    • Obtains P(X = 3) ≈ 0.267.

    Examiner tip: On the exam this is usually read directly from calculator binomial-probability functions, but the underlying formula must still be stated for method marks.

  2. 2.

    Marking analysis: A learner attempts the following task: “X ~ B(10, 0.3). Find P(X = 3).” Their response addresses only this point: “States the binomial probability formula P(X = k) = ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States the binomial probability formula P(X = k) = ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes n = 10, k = 3, p = 0.3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates ¹⁰C₃ = 120. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates (0.3)³(0.7)⁷ ≈ 0.002224. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains P(X = 3) ≈ 0.267. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States the binomial probability formula P(X = k) = ⁿCₖ pᵏ(1 − p)ⁿ⁻ᵏ.
    • Identifies the missing requirement: Substitutes n = 10, k = 3, p = 0.3.
    • Identifies the missing requirement: Calculates ¹⁰C₃ = 120.
    • Identifies the missing requirement: Calculates (0.3)³(0.7)⁷ ≈ 0.002224.
    • Identifies the missing requirement: Obtains P(X = 3) ≈ 0.267.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  3. 3.

    X ~ B(20, 0.25). Find E(X) and Var(X).

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States the formulas E(X) = np and Var(X) = np(1 − p). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes n = 20, p = 0.25 into E(X). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Obtains E(X) = 5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates Var(X) = 20(0.25)(0.75) = 3.75. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: These two formulas apply only to a binomial random variable — check the distribution is genuinely binomial before using them.

    Marking points

    • States the formulas E(X) = np and Var(X) = np(1 − p).
    • Substitutes n = 20, p = 0.25 into E(X).
    • Obtains E(X) = 5.
    • Calculates Var(X) = 20(0.25)(0.75) = 3.75.

    Examiner tip: These two formulas apply only to a binomial random variable — check the distribution is genuinely binomial before using them.

  4. 4.

    Marking analysis: A learner attempts the following task: “X ~ B(20, 0.25). Find E(X) and Var(X).” Their response addresses only this point: “States the formulas E(X) = np and Var(X) = np(1 − p).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States the formulas E(X) = np and Var(X) = np(1 − p). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes n = 20, p = 0.25 into E(X). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains E(X) = 5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates Var(X) = 20(0.25)(0.75) = 3.75. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States the formulas E(X) = np and Var(X) = np(1 − p).
    • Identifies the missing requirement: Substitutes n = 20, p = 0.25 into E(X).
    • Identifies the missing requirement: Obtains E(X) = 5.
    • Identifies the missing requirement: Calculates Var(X) = 20(0.25)(0.75) = 3.75.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  5. 5.

    A discrete random variable X has the probability distribution P(X = 1) = 0.2, P(X = 2) = 0.5, P(X = 3) = 0.3. Find E(X) and Var(X).

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States E(X) = Σx·P(X = x). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Calculates E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States the general variance formula Var(X) = E(X²) − [E(X)]². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates E(X²) = 1²(0.2) + 2²(0.5) + 3²(0.3) = 4.9. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Substitutes to obtain Var(X) = 4.9 − 2.1². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Work through this mathematical step: States Var(X) = 0.49. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This general E(X²) − [E(X)]² route works for any discrete random variable, not just named distributions like the binomial — it's the formula to fall back on whenever no shortcut formula applies.

    Marking points

    • States E(X) = Σx·P(X = x).
    • Calculates E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1.
    • States the general variance formula Var(X) = E(X²) − [E(X)]².
    • Calculates E(X²) = 1²(0.2) + 2²(0.5) + 3²(0.3) = 4.9.
    • Substitutes to obtain Var(X) = 4.9 − 2.1².
    • States Var(X) = 0.49.

    Examiner tip: This general E(X²) − [E(X)]² route works for any discrete random variable, not just named distributions like the binomial — it's the formula to fall back on whenever no shortcut formula applies.

  6. 6.

    Marking analysis: A learner attempts the following task: “A discrete random variable X has the probability distribution P(X = 1) = 0.2, P(X = 2) = 0.5, P(X = 3) = 0.3. Find E(X) and Var(X).” Their response addresses only this point: “States E(X) = Σx·P(X = x).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States E(X) = Σx·P(X = x). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States the general variance formula Var(X) = E(X²) − [E(X)]². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates E(X²) = 1²(0.2) + 2²(0.5) + 3²(0.3) = 4.9. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Substitutes to obtain Var(X) = 4.9 − 2.1². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Requirement 6: Identifies the missing requirement: States Var(X) = 0.49. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States E(X) = Σx·P(X = x).
    • Identifies the missing requirement: Calculates E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1.
    • Identifies the missing requirement: States the general variance formula Var(X) = E(X²) − [E(X)]².
    • Identifies the missing requirement: Calculates E(X²) = 1²(0.2) + 2²(0.5) + 3²(0.3) = 4.9.
    • Identifies the missing requirement: Substitutes to obtain Var(X) = 4.9 − 2.1².
    • Identifies the missing requirement: States Var(X) = 0.49.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  7. 7.

    A continuous random variable X has probability density function f(x) = kx for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. Find the value of k, and hence find P(X > 2).

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that a valid pdf must satisfy ∫₀⁴ f(x) dx = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Integrates: ∫₀⁴ kx dx = k[x²/2]₀⁴ = 8k. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Sets 8k = 1 and solves to obtain k = 1/8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Sets up P(X > 2) = ∫₂⁴ (1/8)x dx. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Integrates to obtain (1/16)[x²]₂⁴ = (1/16)(16 − 4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Work through this mathematical step: States P(X > 2) = 3/4. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Finding the normalising constant k always comes first for a continuous distribution — every subsequent probability calculation depends on it.

    Marking points

    • States that a valid pdf must satisfy ∫₀⁴ f(x) dx = 1.
    • Integrates: ∫₀⁴ kx dx = k[x²/2]₀⁴ = 8k.
    • Sets 8k = 1 and solves to obtain k = 1/8.
    • Sets up P(X > 2) = ∫₂⁴ (1/8)x dx.
    • Integrates to obtain (1/16)[x²]₂⁴ = (1/16)(16 − 4).
    • States P(X > 2) = 3/4.

    Examiner tip: Finding the normalising constant k always comes first for a continuous distribution — every subsequent probability calculation depends on it.

  8. 8.

    Marking analysis: A learner attempts the following task: “A continuous random variable X has probability density function f(x) = kx for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. Find the value of k, and hence find P(X > 2).” Their response addresses only this point: “States that a valid pdf must satisfy ∫₀⁴ f(x) dx = 1.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that a valid pdf must satisfy ∫₀⁴ f(x) dx = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Integrates: ∫₀⁴ kx dx = k[x²/2]₀⁴ = 8k. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Sets 8k = 1 and solves to obtain k = 1/8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Sets up P(X > 2) = ∫₂⁴ (1/8)x dx. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Integrates to obtain (1/16)[x²]₂⁴ = (1/16)(16 − 4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Requirement 6: Identifies the missing requirement: States P(X > 2) = 3/4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that a valid pdf must satisfy ∫₀⁴ f(x) dx = 1.
    • Identifies the missing requirement: Integrates: ∫₀⁴ kx dx = k[x²/2]₀⁴ = 8k.
    • Identifies the missing requirement: Sets 8k = 1 and solves to obtain k = 1/8.
    • Identifies the missing requirement: Sets up P(X > 2) = ∫₂⁴ (1/8)x dx.
    • Identifies the missing requirement: Integrates to obtain (1/16)[x²]₂⁴ = (1/16)(16 − 4).
    • Identifies the missing requirement: States P(X > 2) = 3/4.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  9. 9.

    A continuous random variable X has probability density function f(x) = 3x² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Find E(X).

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States the formula for the expectation of a continuous random variable: E(X) = ∫x f(x) dx over the domain. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets up the integral E(X) = ∫₀¹ x(3x²) dx = ∫₀¹ 3x³ dx. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Integrates to obtain 3[x⁴/4]₀¹. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: States E(X) = 3/4. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: E(X) for a continuous variable replaces the sum Σx·P(X=x) used for discrete variables with the integral ∫x f(x) dx — the underlying idea (a weighted average) is identical.

    Marking points

    • States the formula for the expectation of a continuous random variable: E(X) = ∫x f(x) dx over the domain.
    • Sets up the integral E(X) = ∫₀¹ x(3x²) dx = ∫₀¹ 3x³ dx.
    • Integrates to obtain 3[x⁴/4]₀¹.
    • States E(X) = 3/4.

    Examiner tip: E(X) for a continuous variable replaces the sum Σx·P(X=x) used for discrete variables with the integral ∫x f(x) dx — the underlying idea (a weighted average) is identical.

  10. 10.

    Marking analysis: A learner attempts the following task: “A continuous random variable X has probability density function f(x) = 3x² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Find E(X).” Their response addresses only this point: “States the formula for the expectation of a continuous random variable: E(X) = ∫x f(x) dx over the domain.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States the formula for the expectation of a continuous random variable: E(X) = ∫x f(x) dx over the domain. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets up the integral E(X) = ∫₀¹ x(3x²) dx = ∫₀¹ 3x³ dx. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Integrates to obtain 3[x⁴/4]₀¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: States E(X) = 3/4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States the formula for the expectation of a continuous random variable: E(X) = ∫x f(x) dx over the domain.
    • Identifies the missing requirement: Sets up the integral E(X) = ∫₀¹ x(3x²) dx = ∫₀¹ 3x³ dx.
    • Identifies the missing requirement: Integrates to obtain 3[x⁴/4]₀¹.
    • Identifies the missing requirement: States E(X) = 3/4.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  11. 11.

    A rare disease affects 1% of a population. A test for the disease is 95% accurate for people who have it (true positive) and gives a false positive for 5% of people who do not have it. A person tests positive. Find the probability that they actually have the disease.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States Bayes' theorem: P(D | +) = P(+ | D)P(D) / P(+). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Identifies P(D) = 0.01, P(+ | D) = 0.95, P(+ | D′) = 0.05. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses the law of total probability: P(+) = P(+ | D)P(D) + P(+ | D′)P(D′) = 0.95(0.01) + 0.05(0.99). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates P(+) = 0.0095 + 0.0495 = 0.059. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Substitutes into Bayes' theorem: P(D | +) = 0.0095/0.059. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Work through this mathematical step: States P(D | +) ≈ 0.161 (about 16.1%). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A highly accurate test can still give a low P(D | +) when the disease itself is rare — this counter-intuitive result is exactly why Bayes' theorem matters, and it always requires the law of total probability to find P(+) first.

    Marking points

    • States Bayes' theorem: P(D | +) = P(+ | D)P(D) / P(+).
    • Identifies P(D) = 0.01, P(+ | D) = 0.95, P(+ | D′) = 0.05.
    • Uses the law of total probability: P(+) = P(+ | D)P(D) + P(+ | D′)P(D′) = 0.95(0.01) + 0.05(0.99).
    • Calculates P(+) = 0.0095 + 0.0495 = 0.059.
    • Substitutes into Bayes' theorem: P(D | +) = 0.0095/0.059.
    • States P(D | +) ≈ 0.161 (about 16.1%).

    Examiner tip: A highly accurate test can still give a low P(D | +) when the disease itself is rare — this counter-intuitive result is exactly why Bayes' theorem matters, and it always requires the law of total probability to find P(+) first.

  12. 12.

    Marking analysis: A learner attempts the following task: “A rare disease affects 1% of a population. A test for the disease is 95% accurate for people who have it (true positive) and gives a false positive for 5% of people who do not have it. A person tests positive. Find the probability that they actually have the disease.” Their response addresses only this point: “States Bayes' theorem: P(D | +) = P(+ | D)P(D) / P(+).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States Bayes' theorem: P(D | +) = P(+ | D)P(D) / P(+). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Identifies P(D) = 0.01, P(+ | D) = 0.95, P(+ | D′) = 0.05. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses the law of total probability: P(+) = P(+ | D)P(D) + P(+ | D′)P(D′) = 0.95(0.01) + 0.05(0.99). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates P(+) = 0.0095 + 0.0495 = 0.059. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Substitutes into Bayes' theorem: P(D | +) = 0.0095/0.059. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Requirement 6: Identifies the missing requirement: States P(D | +) ≈ 0.161 (about 16.1%). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States Bayes' theorem: P(D | +) = P(+ | D)P(D) / P(+).
    • Identifies the missing requirement: Identifies P(D) = 0.01, P(+ | D) = 0.95, P(+ | D′) = 0.05.
    • Identifies the missing requirement: Uses the law of total probability: P(+) = P(+ | D)P(D) + P(+ | D′)P(D′) = 0.95(0.01) + 0.05(0.99).
    • Identifies the missing requirement: Calculates P(+) = 0.0095 + 0.0495 = 0.059.
    • Identifies the missing requirement: Substitutes into Bayes' theorem: P(D | +) = 0.0095/0.059.
    • Identifies the missing requirement: States P(D | +) ≈ 0.161 (about 16.1%).

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  13. 13.

    Factory A supplies 60% of a component and Factory B supplies the remaining 40%. 2% of Factory A's components are defective, and 5% of Factory B's components are defective. A component is chosen at random and found to be defective. Find the probability it came from Factory B.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States Bayes' theorem: P(B | D) = P(D | B)P(B) / P(D). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Identifies P(A) = 0.6, P(B) = 0.4, P(D | A) = 0.02, P(D | B) = 0.05. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses the law of total probability: P(D) = P(D | A)P(A) + P(D | B)P(B) = 0.02(0.6) + 0.05(0.4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates P(D) = 0.012 + 0.02 = 0.032. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Substitutes into Bayes' theorem: P(B | D) = 0.02/0.032. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Work through this mathematical step: States P(B | D) = 0.625. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: With more than two suppliers or causes, the denominator P(D) always sums one D-given-cause term per cause — miss a term and every answer that follows is wrong.

    Marking points

    • States Bayes' theorem: P(B | D) = P(D | B)P(B) / P(D).
    • Identifies P(A) = 0.6, P(B) = 0.4, P(D | A) = 0.02, P(D | B) = 0.05.
    • Uses the law of total probability: P(D) = P(D | A)P(A) + P(D | B)P(B) = 0.02(0.6) + 0.05(0.4).
    • Calculates P(D) = 0.012 + 0.02 = 0.032.
    • Substitutes into Bayes' theorem: P(B | D) = 0.02/0.032.
    • States P(B | D) = 0.625.

    Examiner tip: With more than two suppliers or causes, the denominator P(D) always sums one D-given-cause term per cause — miss a term and every answer that follows is wrong.

  14. 14.

    Marking analysis: A learner attempts the following task: “Factory A supplies 60% of a component and Factory B supplies the remaining 40%. 2% of Factory A's components are defective, and 5% of Factory B's components are defective. A component is chosen at random and found to be defective. Find the probability it came from Factory B.” Their response addresses only this point: “States Bayes' theorem: P(B | D) = P(D | B)P(B) / P(D).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [6 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States Bayes' theorem: P(B | D) = P(D | B)P(B) / P(D). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Identifies P(A) = 0.6, P(B) = 0.4, P(D | A) = 0.02, P(D | B) = 0.05. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses the law of total probability: P(D) = P(D | A)P(A) + P(D | B)P(B) = 0.02(0.6) + 0.05(0.4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates P(D) = 0.012 + 0.02 = 0.032. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Substitutes into Bayes' theorem: P(B | D) = 0.02/0.032. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Requirement 6: Identifies the missing requirement: States P(B | D) = 0.625. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States Bayes' theorem: P(B | D) = P(D | B)P(B) / P(D).
    • Identifies the missing requirement: Identifies P(A) = 0.6, P(B) = 0.4, P(D | A) = 0.02, P(D | B) = 0.05.
    • Identifies the missing requirement: Uses the law of total probability: P(D) = P(D | A)P(A) + P(D | B)P(B) = 0.02(0.6) + 0.05(0.4).
    • Identifies the missing requirement: Calculates P(D) = 0.012 + 0.02 = 0.032.
    • Identifies the missing requirement: Substitutes into Bayes' theorem: P(B | D) = 0.02/0.032.
    • Identifies the missing requirement: States P(B | D) = 0.625.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  15. 15.

    A continuous random variable X has probability density function f(x) = x/8 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. Find the median of X.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that the median m satisfies ∫₀ᵐ f(x) dx = 0.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets up the integral ∫₀ᵐ (x/8) dx = 0.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Integrates to obtain [x²/16]₀ᵐ = 0.5, so m²/16 = 0.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Solves to obtain m² = 8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: States the median m = 2√2 (≈ 2.83), rejecting the negative root since X is defined only on [0, 4]. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The median of a continuous random variable is the value that splits the area under the pdf exactly in half — it is generally not the same value as the mean unless the distribution is symmetric.

    Marking points

    • States that the median m satisfies ∫₀ᵐ f(x) dx = 0.5.
    • Sets up the integral ∫₀ᵐ (x/8) dx = 0.5.
    • Integrates to obtain [x²/16]₀ᵐ = 0.5, so m²/16 = 0.5.
    • Solves to obtain m² = 8.
    • States the median m = 2√2 (≈ 2.83), rejecting the negative root since X is defined only on [0, 4].

    Examiner tip: The median of a continuous random variable is the value that splits the area under the pdf exactly in half — it is generally not the same value as the mean unless the distribution is symmetric.

  16. 16.

    Marking analysis: A learner attempts the following task: “A continuous random variable X has probability density function f(x) = x/8 for 0 ≤ x ≤ 4, and f(x) = 0 otherwise. Find the median of X.” Their response addresses only this point: “States that the median m satisfies ∫₀ᵐ f(x) dx = 0.5.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that the median m satisfies ∫₀ᵐ f(x) dx = 0.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets up the integral ∫₀ᵐ (x/8) dx = 0.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Integrates to obtain [x²/16]₀ᵐ = 0.5, so m²/16 = 0.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Solves to obtain m² = 8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: States the median m = 2√2 (≈ 2.83), rejecting the negative root since X is defined only on [0, 4]. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that the median m satisfies ∫₀ᵐ f(x) dx = 0.5.
    • Identifies the missing requirement: Sets up the integral ∫₀ᵐ (x/8) dx = 0.5.
    • Identifies the missing requirement: Integrates to obtain [x²/16]₀ᵐ = 0.5, so m²/16 = 0.5.
    • Identifies the missing requirement: Solves to obtain m² = 8.
    • Identifies the missing requirement: States the median m = 2√2 (≈ 2.83), rejecting the negative root since X is defined only on [0, 4].

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  17. 17.

    The heights of adult males are normally distributed with mean 175 cm and standard deviation 7 cm. Find the probability that a randomly selected male is taller than 185 cm.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States X ~ N(175, 7²). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Standardises: z = (185 − 175)/7 ≈ 1.43. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses the standard normal distribution to find P(Z > 1.43). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains P(X > 185) ≈ 0.0764 (about 7.6%). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: On the IB exam, normal distribution probabilities are read directly from the calculator's built-in distribution functions using the mean and standard deviation — standardising to a z-score is conceptually useful but not strictly required for the final numeric answer.

    Marking points

    • States X ~ N(175, 7²).
    • Standardises: z = (185 − 175)/7 ≈ 1.43.
    • Uses the standard normal distribution to find P(Z > 1.43).
    • Obtains P(X > 185) ≈ 0.0764 (about 7.6%).

    Examiner tip: On the IB exam, normal distribution probabilities are read directly from the calculator's built-in distribution functions using the mean and standard deviation — standardising to a z-score is conceptually useful but not strictly required for the final numeric answer.

  18. 18.

    Marking analysis: A learner attempts the following task: “The heights of adult males are normally distributed with mean 175 cm and standard deviation 7 cm. Find the probability that a randomly selected male is taller than 185 cm.” Their response addresses only this point: “States X ~ N(175, 7²).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States X ~ N(175, 7²). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Standardises: z = (185 − 175)/7 ≈ 1.43. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses the standard normal distribution to find P(Z > 1.43). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains P(X > 185) ≈ 0.0764 (about 7.6%). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States X ~ N(175, 7²).
    • Identifies the missing requirement: Standardises: z = (185 − 175)/7 ≈ 1.43.
    • Identifies the missing requirement: Uses the standard normal distribution to find P(Z > 1.43).
    • Identifies the missing requirement: Obtains P(X > 185) ≈ 0.0764 (about 7.6%).

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  19. 19.

    The number of emails a person receives per hour follows a Poisson distribution with mean 4. Find the probability that exactly 6 emails are received in a given hour.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States X ~ Po(4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: States the Poisson probability formula P(X = k) = e⁻ᵏλᵏ/k! (with λ = 4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Substitutes λ = 4, k = 6: e⁻⁴(4⁶)/6!. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains P(X = 6) ≈ 0.1042. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A Poisson distribution models the number of independent events occurring in a fixed interval at a constant average rate — recognising this scenario (rare events over time or space) is the key to identifying when to use it.

    Marking points

    • States X ~ Po(4).
    • States the Poisson probability formula P(X = k) = e⁻ᵏλᵏ/k! (with λ = 4).
    • Substitutes λ = 4, k = 6: e⁻⁴(4⁶)/6!.
    • Obtains P(X = 6) ≈ 0.1042.

    Examiner tip: A Poisson distribution models the number of independent events occurring in a fixed interval at a constant average rate — recognising this scenario (rare events over time or space) is the key to identifying when to use it.

  20. 20.

    Marking analysis: A learner attempts the following task: “The number of emails a person receives per hour follows a Poisson distribution with mean 4. Find the probability that exactly 6 emails are received in a given hour.” Their response addresses only this point: “States X ~ Po(4).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States X ~ Po(4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States the Poisson probability formula P(X = k) = e⁻ᵏλᵏ/k! (with λ = 4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Substitutes λ = 4, k = 6: e⁻⁴(4⁶)/6!. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains P(X = 6) ≈ 0.1042. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States X ~ Po(4).
    • Identifies the missing requirement: States the Poisson probability formula P(X = k) = e⁻ᵏλᵏ/k! (with λ = 4).
    • Identifies the missing requirement: Substitutes λ = 4, k = 6: e⁻⁴(4⁶)/6!.
    • Identifies the missing requirement: Obtains P(X = 6) ≈ 0.1042.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  21. 21.

    A continuous random variable X has probability density function f(x) = 3x² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Given that E(X) = 3/4, find Var(X).

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States Var(X) = E(X²) − [E(X)]². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets up E(X²) = ∫₀¹ x²(3x²) dx = ∫₀¹ 3x⁴ dx. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Integrates to obtain E(X²) = 3[x⁵/5]₀¹ = 3/5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Substitutes Var(X) = 3/5 − (3/4)². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains Var(X) = 3/80 (= 0.0375). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Var(X) = E(X²) − [E(X)]² is the same general formula used for discrete random variables, with the sum replaced by an integral — always find E(X²) using the pdf before subtracting the squared mean.

    Marking points

    • States Var(X) = E(X²) − [E(X)]².
    • Sets up E(X²) = ∫₀¹ x²(3x²) dx = ∫₀¹ 3x⁴ dx.
    • Integrates to obtain E(X²) = 3[x⁵/5]₀¹ = 3/5.
    • Substitutes Var(X) = 3/5 − (3/4)².
    • Obtains Var(X) = 3/80 (= 0.0375).

    Examiner tip: Var(X) = E(X²) − [E(X)]² is the same general formula used for discrete random variables, with the sum replaced by an integral — always find E(X²) using the pdf before subtracting the squared mean.

  22. 22.

    Marking analysis: A learner attempts the following task: “A continuous random variable X has probability density function f(x) = 3x² for 0 ≤ x ≤ 1, and f(x) = 0 otherwise. Given that E(X) = 3/4, find Var(X).” Their response addresses only this point: “States Var(X) = E(X²) − [E(X)]².” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States Var(X) = E(X²) − [E(X)]². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets up E(X²) = ∫₀¹ x²(3x²) dx = ∫₀¹ 3x⁴ dx. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Integrates to obtain E(X²) = 3[x⁵/5]₀¹ = 3/5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Substitutes Var(X) = 3/5 − (3/4)². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains Var(X) = 3/80 (= 0.0375). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States Var(X) = E(X²) − [E(X)]².
    • Identifies the missing requirement: Sets up E(X²) = ∫₀¹ x²(3x²) dx = ∫₀¹ 3x⁴ dx.
    • Identifies the missing requirement: Integrates to obtain E(X²) = 3[x⁵/5]₀¹ = 3/5.
    • Identifies the missing requirement: Substitutes Var(X) = 3/5 − (3/4)².
    • Identifies the missing requirement: Obtains Var(X) = 3/80 (= 0.0375).

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  23. 23.

    For two events A and B, P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2. Find P(A | B).

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States P(A | B) = P(A ∩ B)/P(B). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes 0.2/0.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Obtains P(A | B) = 0.4. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Conditional probability restricts attention to only the outcomes where B has already occurred, then asks what fraction of those also satisfy A — this is exactly what dividing by P(B) achieves.

    Marking points

    • States P(A | B) = P(A ∩ B)/P(B).
    • Substitutes 0.2/0.5.
    • Obtains P(A | B) = 0.4.

    Examiner tip: Conditional probability restricts attention to only the outcomes where B has already occurred, then asks what fraction of those also satisfy A — this is exactly what dividing by P(B) achieves.

  24. 24.

    Marking analysis: A learner attempts the following task: “For two events A and B, P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2. Find P(A | B).” Their response addresses only this point: “States P(A | B) = P(A ∩ B)/P(B).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States P(A | B) = P(A ∩ B)/P(B). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes 0.2/0.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains P(A | B) = 0.4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States P(A | B) = P(A ∩ B)/P(B).
    • Identifies the missing requirement: Substitutes 0.2/0.5.
    • Identifies the missing requirement: Obtains P(A | B) = 0.4.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  25. 25.

    X and Y are independent random variables with E(X) = 5, E(Y) = 3, Var(X) = 4 and Var(Y) = 9. Find E(X + Y) and Var(X + Y).

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that E(X + Y) = E(X) + E(Y), which holds regardless of independence. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Obtains E(X + Y) = 5 + 3 = 8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States that, since X and Y are independent, Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The expectation of a sum always adds directly, whether or not the variables are independent — but the variance of a sum only adds directly when the variables are independent, since otherwise a covariance term must be included.

    Marking points

    • States that E(X + Y) = E(X) + E(Y), which holds regardless of independence.
    • Obtains E(X + Y) = 5 + 3 = 8.
    • States that, since X and Y are independent, Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13.

    Examiner tip: The expectation of a sum always adds directly, whether or not the variables are independent — but the variance of a sum only adds directly when the variables are independent, since otherwise a covariance term must be included.

  26. 26.

    Marking analysis: A learner attempts the following task: “X and Y are independent random variables with E(X) = 5, E(Y) = 3, Var(X) = 4 and Var(Y) = 9. Find E(X + Y) and Var(X + Y).” Their response addresses only this point: “States that E(X + Y) = E(X) + E(Y), which holds regardless of independence.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that E(X + Y) = E(X) + E(Y), which holds regardless of independence. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Obtains E(X + Y) = 5 + 3 = 8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that, since X and Y are independent, Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that E(X + Y) = E(X) + E(Y), which holds regardless of independence.
    • Identifies the missing requirement: Obtains E(X + Y) = 5 + 3 = 8.
    • Identifies the missing requirement: States that, since X and Y are independent, Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  27. 27.

    X is a random variable with E(X) = 10 and Var(X) = 4. Find E(3X − 2) and Var(3X − 2).

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States E(aX + b) = aE(X) + b. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Obtains E(3X − 2) = 3(10) − 2 = 28. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States Var(aX + b) = a²Var(X), obtaining Var(3X − 2) = 9(4) = 36. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Adding a constant b shifts the expectation but never affects the variance (since it does not change the spread) — only the multiplicative factor a affects variance, and it does so by a factor of a², not a.

    Marking points

    • States E(aX + b) = aE(X) + b.
    • Obtains E(3X − 2) = 3(10) − 2 = 28.
    • States Var(aX + b) = a²Var(X), obtaining Var(3X − 2) = 9(4) = 36.

    Examiner tip: Adding a constant b shifts the expectation but never affects the variance (since it does not change the spread) — only the multiplicative factor a affects variance, and it does so by a factor of a², not a.

  28. 28.

    Marking analysis: A learner attempts the following task: “X is a random variable with E(X) = 10 and Var(X) = 4. Find E(3X − 2) and Var(3X − 2).” Their response addresses only this point: “States E(aX + b) = aE(X) + b.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States E(aX + b) = aE(X) + b. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Obtains E(3X − 2) = 3(10) − 2 = 28. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States Var(aX + b) = a²Var(X), obtaining Var(3X − 2) = 9(4) = 36. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States E(aX + b) = aE(X) + b.
    • Identifies the missing requirement: Obtains E(3X − 2) = 3(10) − 2 = 28.
    • Identifies the missing requirement: States Var(aX + b) = a²Var(X), obtaining Var(3X − 2) = 9(4) = 36.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  29. 29.

    A hand of 5 cards is dealt from a standard deck of 52 cards. Find the probability that the hand contains exactly 3 aces.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States the total number of possible 5-card hands: ⁵²C₅. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: States the number of ways to choose 3 aces from the 4 available: ⁴C₃. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States the number of ways to choose the remaining 2 cards from the 48 non-ace cards: ⁴⁸C₂. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates the probability as (⁴C₃ × ⁴⁸C₂)/⁵²C₅ = (4 × 1128)/2 598 960. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains a probability of approximately 0.00174 (about 0.174%). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This 'choose separately, then multiply' structure works whenever a selection requires a fixed number of items from two or more distinct groups (here: aces and non-aces) — always divide by the total number of ways to choose the full hand.

    Marking points

    • States the total number of possible 5-card hands: ⁵²C₅.
    • States the number of ways to choose 3 aces from the 4 available: ⁴C₃.
    • States the number of ways to choose the remaining 2 cards from the 48 non-ace cards: ⁴⁸C₂.
    • Calculates the probability as (⁴C₃ × ⁴⁸C₂)/⁵²C₅ = (4 × 1128)/2 598 960.
    • Obtains a probability of approximately 0.00174 (about 0.174%).

    Examiner tip: This 'choose separately, then multiply' structure works whenever a selection requires a fixed number of items from two or more distinct groups (here: aces and non-aces) — always divide by the total number of ways to choose the full hand.

  30. 30.

    Marking analysis: A learner attempts the following task: “A hand of 5 cards is dealt from a standard deck of 52 cards. Find the probability that the hand contains exactly 3 aces.” Their response addresses only this point: “States the total number of possible 5-card hands: ⁵²C₅.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States the total number of possible 5-card hands: ⁵²C₅. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States the number of ways to choose 3 aces from the 4 available: ⁴C₃. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States the number of ways to choose the remaining 2 cards from the 48 non-ace cards: ⁴⁸C₂. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates the probability as (⁴C₃ × ⁴⁸C₂)/⁵²C₅ = (4 × 1128)/2 598 960. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains a probability of approximately 0.00174 (about 0.174%). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States the total number of possible 5-card hands: ⁵²C₅.
    • Identifies the missing requirement: States the number of ways to choose 3 aces from the 4 available: ⁴C₃.
    • Identifies the missing requirement: States the number of ways to choose the remaining 2 cards from the 48 non-ace cards: ⁴⁸C₂.
    • Identifies the missing requirement: Calculates the probability as (⁴C₃ × ⁴⁸C₂)/⁵²C₅ = (4 × 1128)/2 598 960.
    • Identifies the missing requirement: Obtains a probability of approximately 0.00174 (about 0.174%).

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.