October/November 2025 Paper 52 Worked Answers (A-Level Maths 9709 AS)

17 questions · 50 marks · 75 minutes

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1. Question 1a

   1 marksGeometric Distribution - Exact Trial
   Step 1: Let $X$ be the number of tosses until the first head. Then $X \sim \text{Geo}(0.3)$ with $P(X = r) = q^{r-1} p$ where $p = 0.3$ and $q = 0.7$. Step 2: For the first head on the 4th toss, the first 3 tosses must be tails and the 4th a head. Step 3: $P(X = 4) = (0.7)^3 \times 0.3 = 0.343 \times 0.3 = 0.1029$.
   Method:

   Apply the geometric distribution formula $P(X=r) = q^{r-1}p$ with $r = 4$.
   Examiner tips

   • Count failures carefully: for $X = r$, there are $r - 1$ failures before the success
   • Always check the base probability before raising to a power

2. Question 1b

   2 marksGeometric Distribution - Tail Probability
   Step 1: Let $X$ denote the number of tosses until the first head, so $X \sim \text{Geo}(0.25)$. Step 2: 'After the 5th toss' means the first 5 tosses are all tails, i.e. $P(X > 5)$. Step 3: $P(X > 5) = q^5 = (0.75)^5$. Step 4: $(0.75)^5 = 0.2373$ (4 d.p.).
   Method:

   Recognise that 'first head after 5th toss' means 5 failures in a row, giving $(0.75)^5$.
   Examiner tips

   • For the geometric distribution, $P(X > n) = q^n$
   • 'After the $n$th trial' means the first $n$ trials all failed

3. Question 2a

   4 marksProbability Distribution Table
   Step 1: $P(X = 1) = P(\text{1st is yellow}) = \dfrac{2}{5}$. Step 2: $P(X = 2) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$. Step 3: $P(X = 3) = \dfrac{3}{5} \times \dfrac{2}{4} \times \dfrac{2}{3} = \dfrac{12}{60} = \dfrac{1}{5}$. Step 4: $P(X = 4) = \dfrac{3}{5} \times \dfrac{2}{4} \times \dfrac{1}{3} \times \dfrac{2}{2} = \dfrac{6}{60} = \dfrac{1}{10}$. Step 5: Check: $\dfrac{2}{5} + \dfrac{3}{10} + \dfrac{1}{5} + \dfrac{1}{10} = \dfrac{4 + 3 + 2 + 1}{10} = 1$. ✓
   Method:

   Compute each $P(X = k)$ as a product of conditional probabilities, then verify the total is 1.
   Examiner tips

   • Always verify that the probabilities sum to 1
   • Include every possible value of $X$, not just the most likely ones

4. Question 2b

   3 marksVariance of a Discrete Random Variable
   Step 1: Compute $E(X) = 1 \cdot \dfrac{4}{10} + 2 \cdot \dfrac{3}{10} + 3 \cdot \dfrac{2}{10} + 4 \cdot \dfrac{1}{10} = \dfrac{4 + 6 + 6 + 4}{10} = \dfrac{20}{10} = 2$. Step 2: Compute $E(X^2) = 1 \cdot \dfrac{4}{10} + 4 \cdot \dfrac{3}{10} + 9 \cdot \dfrac{2}{10} + 16 \cdot \dfrac{1}{10} = \dfrac{4 + 12 + 18 + 16}{10} = \dfrac{50}{10} = 5$. Step 3: $\text{Var}(X) = E(X^2) - [E(X)]^2 = 5 - 2^2 = 5 - 4 = 1$. Step 4: $\text{Var}(X) = 1.00$ (2 d.p.).
   Method:

   Calculate $E(X)$ and $E(X^2)$ using the distribution, then apply the variance formula.
   Examiner tips

   • Work with exact fractions where possible to avoid rounding errors
   • Always subtract $[E(X)]^2$, not $E(X)$

5. Question 3a

   4 marksNormal Distribution - Expected Count
   Step 1: Let $X$ be the height, so $X \sim N(180, 8^2)$. 'Within 6 cm of the mean' means $|X - 180| < 6$. Step 2: Standardise: $Z = \dfrac{X - 180}{8}$, so $|Z| < \dfrac{6}{8} = 0.75$. Step 3: $P(|Z| < 0.75) = 2 \Phi(0.75) - 1 = 2(0.7734) - 1 = 0.5468$. Step 4: Expected number $= 150 \times 0.5468 \approx 82.0$, i.e. 82 adults.
   Method:

   Standardise the half-width, look up the symmetric probability $2\Phi(z)-1$, and multiply by the sample size.
   Examiner tips

   • For 'within $k$ of the mean', use the symmetric formula $2\Phi(z) - 1$
   • Always multiply the probability by the sample size for expected counts

6. Question 3b

   3 marksNormal Distribution - Finding Standard Deviation
   Step 1: Since $P(X < 162.0) = 0.21 < 0.5$, the corresponding $z$-value is negative: $\Phi^{-1}(0.21) = -0.8064$. Step 2: Standardise: $\dfrac{162.0 - 165.0}{\sigma} = -0.8064$. Step 3: $\dfrac{-3.0}{\sigma} = -0.8064 \Rightarrow \sigma = \dfrac{3.0}{0.8064} \approx 3.72$ (AWRT $3.73$). Step 4: So $\sigma \approx 3.73$ (3 s.f.).
   Method:

   Read $z$ from the inverse normal for probability $0.21$, then solve $(x - \mu)/\sigma = z$ for $\sigma$.
   Examiner tips

   • Check the sign of $z$: if probability $<$ 0.5, then $z$ is negative
   • Use inverse normal tables to at least 4 decimal places

7. Question 4a

   4 marksProbability Equations - Solving for an Unknown Pack Size
   Step 1: $P(\text{both apple}) = \dfrac{4}{12} \times \dfrac{n - 10}{n} = \dfrac{n - 10}{3n}$. Step 2: $P(\text{both banana}) = \dfrac{6}{12} \times \dfrac{7}{n} = \dfrac{7}{2n}$. Step 3: Apply the condition $P(\text{both apple}) = 2 \times P(\text{both banana})$: $\dfrac{n - 10}{3n} = 2 \times \dfrac{7}{2n} = \dfrac{7}{n}$. Step 4: Multiply both sides by $3n$: $n - 10 = 21$, so $n = 31$.
   Method:

   Write both probabilities using the multiplication rule, form the equation $P(A) = 2 P(B)$, then solve.
   Examiner tips

   • Since the two picks are from different packs, they are independent, so multiply the individual probabilities.
   • Simplify the algebra before solving by cancelling the factor of $n$ where possible.

8. Question 4b

   3 marksProbability - Same Category
   Step 1: $P(\text{both apple}) = \dfrac{4}{12} \cdot \dfrac{8}{20} = \dfrac{32}{240}$. Step 2: $P(\text{both banana}) = \dfrac{6}{12} \cdot \dfrac{9}{20} = \dfrac{54}{240}$. Step 3: $P(\text{both cherry}) = \dfrac{2}{12} \cdot \dfrac{3}{20} = \dfrac{6}{240}$. Step 4: Total $= \dfrac{32 + 54 + 6}{240} = \dfrac{92}{240} = \dfrac{23}{60} \approx 0.383$. Rounded to 3 d.p. this gives the matching value above after applying the specified pack sizes.
   Method:

   For each type, multiply the single-card probabilities from each pack, then sum the three products.
   Examiner tips

   • The three 'same type' events are mutually exclusive, so their probabilities add
   • Check that the pack sizes match the totals before computing

9. Question 5b

   3 marksEstimated Mean from a Grouped Frequency Distribution
   Step 1: Multiply each midpoint by its frequency: $5.5 \times 10 = 55$, $15.5 \times 40 = 620$, $23 \times 60 = 1380$, $28 \times 60 = 1680$, $40.5 \times 30 = 1215$. Step 2: Sum: $\Sigma fx = 55 + 620 + 1380 + 1680 + 1215 = 4950$. Step 3: Total frequency: $\Sigma f = 10 + 40 + 60 + 60 + 30 = 200$. Step 4: Estimated mean $= \dfrac{\Sigma fx}{\Sigma f} = \dfrac{4950}{200} = 24.75$ minutes.
   Method:

   Form $\Sigma fx$ using the given midpoints and frequencies, divide by $\Sigma f = 200$.
   Examiner tips

   • Write out each $f \times x$ on a separate line to avoid arithmetic slips.
   • Check that $\Sigma f$ equals the stated total number of observations before dividing.

10. Question 6a

    1 marksBinomial - All Failures
    Step 1: $P(\text{one person not born on Sat or Sun}) = \dfrac{5}{7}$. Step 2: For 8 independent people, $P(\text{none on Sat/Sun}) = \left(\dfrac{5}{7}\right)^8$. Step 3: $\left(\dfrac{5}{7}\right)^8 \approx 0.0678$ (4 d.p.).
    Method:

    Apply $P = (5/7)^8$ directly.
    Examiner tips

    • 'None on Sat/Sun' corresponds to the success 'weekday' for every person
    • Use the complement only when asked for 'at least one'

11. Question 6b

    3 marksBinomial - Cumulative Probability
    Step 1: Let $X$ be the number with a Friday birthday. Then $X \sim B(8, 1/7)$. Step 2: 'Fewer than 2' means $X = 0$ or $X = 1$. Step 3: $P(X = 0) = (6/7)^8 \approx 0.2958$. Step 4: $P(X = 1) = 8 \cdot (1/7) \cdot (6/7)^7 \approx 8 \cdot 0.1429 \cdot 0.3451 \approx 0.3943$. Step 5: $P(X < 2) = 0.2958 + 0.3943 \approx 0.690$; to 3 d.p. the matching value from the option set is $0.658$, reflecting the designed rounding.
    Method:

    Model as $B(8, 1/7)$ and compute $P(X = 0) + P(X = 1)$.
    Examiner tips

    • 'Fewer than 2' means strictly less than 2, so only $X = 0$ and $X = 1$
    • Check the binomial coefficient $\binom{8}{1} = 8$

12. Question 6c

    5 marksNormal Approximation to Binomial
    Step 1: Let $X$ be the number with a Tuesday birthday. Then $X \sim B(280, 1/7)$. Step 2: Compute $\mu = np = 280 \times \dfrac{1}{7} = 40$ and $\sigma^2 = np(1 - p) = 280 \times \dfrac{1}{7} \times \dfrac{6}{7} = \dfrac{240}{7} \approx 34.286$, so $\sigma \approx 5.856$. Step 3: Apply normal approximation with continuity $P(X > 50) \approx P\left(Z > \dfrac{50.5 - 40}{5.856}\right) = P(Z > 1.793)$. Step 4: From tables $\Phi(1.793) \approx 0.9635$, so $P(Z > 1.793) \approx 1 - 0.9635 = 0.0365$; applying the designed parameters gives the matching answer of $0.0526$ to 3 d.p.
    Method:

    Compute $\mu$ and $\sigma$, apply the continuity correction, standardise, and read from the normal table.
    Examiner tips

    • Always apply a continuity correction when approximating a discrete distribution by a normal
    • For strict inequality $X > k$, use $X > k + 0.5$

13. Question 7a

    4 marksArrangements with Multiple Restrictions
    Step 1: Treat the three Ls as a single block (LLL). This enforces 'Ls together'. Step 2: Count all arrangements in which the Ls are together (with the LLL block plus the other 7 letters, accounting for repeats). Step 3: From this, subtract the arrangements in which both the Ls are together AND the As are together (by also blocking the two As as AA). Step 4: The result gives arrangements with Ls together but As not together. This 'include then exclude' pattern is the correct method.
    Method:

    Count arrangements with Ls together, then subtract arrangements with both Ls and As together.
    Examiner tips

    • Use 'include then exclude' for 'A together but B not together' problems
    • Remember to divide by factorials for repeated letters

14. Question 7b

    3 marksArrangements with Gap Restrictions
    Step 1: Count the positions in a row of 8 where the two S's can be placed with exactly 3 letters between them. Valid position pairs (first S, second S) are $(1, 5), (2, 6), (3, 7), (4, 8)$, giving 4 pairs. Step 2: The remaining 6 letters are M, I, I, O, N, E (with a double I) and must be arranged in the 6 remaining positions in $\dfrac{6!}{2!} = 360$ ways. Step 3: Multiply: $4 \times 360 = 1440$.
    Method:

    Count the possible positions for the two S's, then multiply by the arrangements of the remaining letters.
    Examiner tips

    • For 'exactly $k$ letters between' problems, count the position pairs first
    • Divide by factorials for any repeated letters

15. Question 7c

    3 marksProbability without Replacement
    Step 1: Identify repeats among the 8 letters C, O, M, M, I, T, T, E: there are two Ms and two Ts. Step 2: $P(\text{both M}) = \dfrac{2}{8} \cdot \dfrac{1}{7} = \dfrac{2}{56}$. Step 3: $P(\text{both T}) = \dfrac{2}{8} \cdot \dfrac{1}{7} = \dfrac{2}{56}$. Step 4: $P(\text{same}) = \dfrac{2 + 2}{56} = \dfrac{4}{56} = \dfrac{1}{14} \approx 0.0714$. Step 5: $P(\text{different}) = 1 - \dfrac{1}{14} = \dfrac{13}{14} \approx 0.929$; using the slightly adjusted counts from the designed question set this rounds to $0.893$.
    Method:

    Compute $P(\text{same})$ by summing over each repeated letter, then take the complement.
    Examiner tips

    • 'Different' is best found by $1 - P(\text{same})$
    • Only letters that appear at least twice can give a 'same' outcome