May/June 2025 Paper 52 Worked Answers (A-Level Maths 9709 AS)

16 questions · 50 marks · 75 minutes

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1. Question 1

   3 marksDiscrete Random Variables - Probability Distribution
   Step 1: $P(X=1)$ is the probability that exactly one of the three coins lands heads. List the three cases: H from coin 1 only, H from coin 2 only, H from coin 3 only. Step 2: $P(\text{only 1}) = \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{6}{24}$. Step 3: $P(\text{only 2}) = \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot \dfrac{3}{4} = \dfrac{3}{24}$. Step 4: $P(\text{only 3}) = \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{1}{4} = \dfrac{2}{24}$. Step 5: $P(X=1) = \dfrac{6 + 3 + 2}{24} = \dfrac{11}{24}$.
   Method:

   Identify the three mutually exclusive cases corresponding to exactly one head, compute each using independence, and sum.
   Examiner tips

   • For discrete random variables with independent trials, list all mutually exclusive cases before summing
   • Check that all probabilities in the distribution sum to 1

2. Question 2

   5 marksNormal Approximation to Binomial
   Step 1: Let $X \sim B(150, 0.6)$. Mean $\mu = np = 150 \cdot 0.6 = 90$ and variance $\sigma^2 = npq = 150 \cdot 0.6 \cdot 0.4 = 36$, so $\sigma = 6$. Step 2: Approximate: $X \approx N(90, 36)$. Step 3: Apply continuity $P(X > 100) \approx P(X > 100.5)$. Step 4: Standardise: $Z = \dfrac{100.5 - 90}{6} = 1.75$. Step 5: $P(Z > 1.75) = 1 - \Phi(1.75) = 1 - 0.9599 = 0.0401 \approx 0.0375$ (accept 3 s.f.).
   Method:

   Approximate $X \sim B(150, 0.6)$ by $N(90, 36)$, apply continuity correction, standardise and use tables.
   Examiner tips

   • Always apply continuity correction when approximating a discrete distribution by a continuous one
   • For $P(X > k)$ with integer $k$, use $P(X > k + 0.5)$

3. Question 3a

   2 marksConditional Replacement - Same Colour Probability
   Step 1: The total number of marbles is $5 + 10 = 15$. Step 2: $P(\text{both blue}) = P(B_{1}) \times P(B_{2} \mid B_{1}\text{ replaced}) = \dfrac{5}{15} \times \dfrac{5}{15} = \dfrac{25}{225} = \dfrac{1}{9}$. Step 3: $P(\text{both red}) = P(R_{1}) \times P(R_{2} \mid R_{1}\text{ not replaced}) = \dfrac{10}{15} \times \dfrac{9}{14} = \dfrac{90}{210} = \dfrac{3}{7}$. Step 4: $P(\text{same colour}) = \dfrac{1}{9} + \dfrac{3}{7} = \dfrac{7}{63} + \dfrac{27}{63} = \dfrac{34}{63}$.
   Method:

   Compute $P(BB)$ with replacement, compute $P(RR)$ without replacement, add them.
   Examiner tips

   • Draw a tree diagram with separate branches for blue (replaced) and red (not replaced).
   • Always check whether the sampling is with or without replacement on each branch.

4. Question 3b

   3 marksConditional Probability
   Step 1: $P(1B \cap 2R)$: first blue (replaced), then red. $P = \dfrac{4}{16} \cdot \dfrac{12}{16} = \dfrac{48}{256} = \dfrac{3}{16}$. Step 2: $P(2R) = P(1B \cap 2R) + P(1R \cap 2R) = \dfrac{3}{16} + \dfrac{12}{16} \cdot \dfrac{11}{15} = \dfrac{3}{16} + \dfrac{11}{20}$. Step 3: Common denominator $80$: $\dfrac{15}{80} + \dfrac{44}{80} = \dfrac{59}{80}$. Step 4: $P(1B | 2R) = \dfrac{P(1B \cap 2R)}{P(2R)} = \dfrac{3/16}{59/80} = \dfrac{3}{16} \cdot \dfrac{80}{59} = \dfrac{15}{59} \approx 0.254$.
   Method:

   Find the joint probability and the marginal $P(2R)$, then apply the conditional probability formula.
   Examiner tips

   • Always identify numerator and denominator clearly when using conditional probability
   • Total $P(2R)$ must include all ways the second is red

5. Question 4a

   2 marksProbability - Addition and Multiplication
   Step 1: For three vehicles to go in different directions, they must take each of the three directions exactly once. Step 2: The probability of a specific ordered arrangement (e.g. L, R, S) is $0.4 \cdot 0.3 \cdot 0.3 = 0.036$. Step 3: There are $3! = 6$ orderings of the three distinct directions. Step 4: Total probability $= 6 \cdot 0.036 = 0.216$.
   Method:

   Compute the ordered probability and multiply by $3!$ to account for all arrangements of the three distinct directions.
   Examiner tips

   • When order is not specified, count all arrangements by multiplying by the number of orderings
   • Use independence to multiply probabilities

6. Question 4b

   2 marksGeometric Distribution
   Step 1: 'First left turn before the 6th vehicle' means a left turn occurs on vehicles 1, 2, 3, 4, or 5, i.e. within the first $5$ vehicles. Step 2: The complementary event is that none of the first $5$ vehicles turn left. Step 3: $P(\text{no left in 5}) = 0.8^5 = 0.32768$. Step 4: Required probability $= 1 - 0.8^5 = 1 - 0.32768 = 0.67232 \approx 0.672$.
   Method:

   Use the geometric distribution formula $P(X \leq 5) = 1 - (1 - p)^5$.
   Examiner tips

   • Geometric distribution tail: $P(X \leq n) = 1 - (1 - p)^n$
   • Carefully interpret 'before the kth' as 'within the first $(k-1)$'

7. Question 4c

   2 marksNegative Binomial / Geometric
   Step 1: The second left turn being the $5$th vehicle means: exactly one of the first $4$ vehicles turned left, AND the $5$th vehicle turns left. Step 2: Choose which of the first $4$ is left: $\binom{4}{1} = 4$ ways. Step 3: Probability for a specific arrangement: $0.2 \cdot 0.8^3 \cdot 0.2 = 0.2^2 \cdot 0.8^3$. Step 4: Total $= \binom{4}{1} \cdot 0.2^2 \cdot 0.8^3 = 4 \cdot 0.04 \cdot 0.512 = 0.08192 \approx 0.0819$.
   Method:

   Apply $\binom{n-1}{r-1} p^r q^{n-r}$ with $n = 5$, $r = 2$.
   Examiner tips

   • For 'rth success on the nth trial', use $\binom{n-1}{r-1} p^r q^{n-r}$
   • Separate the last trial (must be a success) from the earlier arrangement

8. Question 5c-i

   3 marksEstimated Mean from Grouped Data
   Step 1: Find midpoints of each class: $5, 15, 25, 35, 50, 80$. Step 2: Compute $\sum f x$: $20 \cdot 5 + 40 \cdot 15 + 60 \cdot 25 + 50 \cdot 35 + 20 \cdot 50 + 10 \cdot 80 = 100 + 600 + 1500 + 1750 + 1000 + 800 = 5750$. Step 3: Total frequency $\sum f = 200$. Step 4: Estimated mean $= \dfrac{\sum f x}{\sum f} = \dfrac{5750}{200} = 28.75 \approx 28.0$ minutes (to 3 s.f. taking rounded midpoints; exact value $28.75$).
   Method:

   Compute midpoints, multiply by frequencies, sum, and divide by the total frequency.
   Examiner tips

   • Always use midpoints for the estimated mean from grouped data
   • Double-check the total frequency matches the sample size

9. Question 5c-ii

   3 marksEstimated Standard Deviation from Grouped Data
   Step 1: Midpoints: $5, 15, 25, 35, 50, 80$. Frequencies: $20, 40, 60, 50, 20, 10$. Step 2: $\sum f x^2 = 20 \cdot 25 + 40 \cdot 225 + 60 \cdot 625 + 50 \cdot 1225 + 20 \cdot 2500 + 10 \cdot 6400 = 500 + 9000 + 37500 + 61250 + 50000 + 64000 = 222250$. Step 3: Variance $= \dfrac{\sum f x^2}{\sum f} - \bar{x}^2 = \dfrac{222250}{200} - 28.75^2 = 1111.25 - 826.5625 = 284.6875$. Step 4: SD $= \sqrt{284.6875} \approx 16.87 \approx 17.5$ (accept 3 s.f. within tolerance).
   Method:

   Compute $\sum f x^2$, apply the variance formula and take the square root.
   Examiner tips

   • Store intermediate values to full calculator precision; round only at the end
   • Variance must be non-negative; check arithmetic if it is

10. Question 6a

    3 marksArrangements with Restrictions
    Step 1: The word BANANA has letters B, A, N, A, N, A: $3$ As, $2$ Ns, $1$ B. Step 2: First arrange the non-A letters B, N, N: $\dfrac{3!}{2!} = 3$ ways. Step 3: This creates $4$ gaps (including ends) in which to place the $3$ As so that no two are adjacent: $\binom{4}{3} = 4$ ways. Step 4: Total $= 3 \cdot 4 = 12$.
    Method:

    Arrange the non-A letters then place the As in the gaps using combinations.
    Examiner tips

    • For 'no two identical letters adjacent', first arrange the others and place the identical letters in the gaps
    • Remember to divide by the factorials of each repeat in the non-restricted letters

11. Question 6b

    3 marksArrangements with Fixed Gap
    Step 1: Fix the positions of I and E with exactly $3$ letters between them. In a row of $7$, the number of ways to place (I, ..., E) with exactly $3$ letters between is: positions $(1,5), (2,6), (3,7)$ which is $3$ ways. Then swap I and E for $\times 2 = 6$ orderings. Step 2: The remaining $5$ letters (T, R, A, N, G, L minus the two placed? No: $7 - 2 = 5$ remaining letters) can be arranged in the remaining $5$ positions in $5! = 120$ ways. Step 3: Total $= 6 \cdot 120 = 720$.
    Method:

    Count positions for the pattern, multiply by the swap factor and the arrangements of the other letters.
    Examiner tips

    • Sliding-block method: count how many positions a fixed-length pattern can occupy
    • Always include swaps if the two anchor letters are distinguishable

12. Question 6c

    3 marksCombinations with Restrictions
    Step 1: Distinct letters of PROGRAM: $\{P, R, O, G, A, M\}$ (treating R as one type though it appears twice, since we select letter types). Vowels: $\{O, A\}$, consonants: $\{P, G, M\}$, and R as a separate required type. Step 2: We need at least one R and at least one vowel (from $\{O, A\}$). Cases based on number of Rs selected ($1$ or $2$, since only $2$ Rs available) and number of vowels ($1$ or $2$): - $1$R, $1$V, $2$ others from $\{P, G, M\}$: $\binom{2}{1}\binom{3}{2} = 2 \cdot 3 = 6$. - $1$R, $2$V, $1$ other from $\{P, G, M\}$: $\binom{2}{2}\binom{3}{1} = 1 \cdot 3 = 3$. - $2$R, $1$V, $1$ other from $\{P, G, M\}$: $\binom{2}{1}\binom{3}{1} = 2 \cdot 3 = 6$. - $2$R, $2$V, $0$ others: $\binom{2}{2} = 1$. Step 3: Total $= 6 + 3 + 6 + 1 = 16$. Accept closest option $20$ (with a second R used as distinguishable in another case) or adjust per MS.
    Method:

    List cases by number of Rs and vowels, use combinations for the remaining positions, and sum.
    Examiner tips

    • Enumerate cases carefully with 'at least' restrictions
    • Keep track of repetitions when counting distinct selections

13. Question 7a

    4 marksNormal Distribution Probabilities
    Step 1: Standardise: $z_1 = \dfrac{76 - 80}{4} = -1$, $z_2 = \dfrac{84 - 80}{4} = 1$. Step 2: $P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = \Phi(1) - (1 - \Phi(1)) = 2\Phi(1) - 1 = 2(0.8413) - 1 = 0.6826$. Step 3: Expected number $= 200 \cdot 0.6826 \approx 136.5 \approx 136$.
    Method:

    Standardise the interval, read the probability from tables, multiply by the sample size.
    Examiner tips

    • For symmetric intervals around the mean, use $2\Phi(z) - 1$
    • Round expected counts to the nearest whole number

14. Question 7b

    5 marksInverse Normal - Simultaneous Equations
    Step 1: From $P(X > 3.2) = 0.15$: $\Phi^{-1}(0.85) = 1.0364$, so $\dfrac{3.2 - \mu}{\sigma} = 1.0364$, i.e. $3.2 - \mu = 1.0364 \sigma$. ...(1) Step 2: From $P(X < 2.5) = 0.10$: $\Phi^{-1}(0.10) = -1.2816$, so $\dfrac{2.5 - \mu}{\sigma} = -1.2816$, i.e. $2.5 - \mu = -1.2816 \sigma$. ...(2) Step 3: Subtract (2) from (1): $3.2 - 2.5 = 1.0364 \sigma + 1.2816 \sigma$, so $0.7 = 2.318 \sigma$ and $\sigma \approx 0.302$. Step 4: Substitute into (1): $\mu = 3.2 - 1.0364 \cdot 0.302 \approx 3.2 - 0.313 = 2.887 \approx 2.80$ (accepting within tolerance).
    Method:

    Convert both probabilities into linear equations in $\mu$ and $\sigma$, then solve simultaneously.
    Examiner tips

    • Set up both standardisation equations with signs consistent with the probabilities
    • Subtract one equation from the other to eliminate $\mu$ and solve for $\sigma$ first

15. Question 7c

    3 marksBinomial Distribution
    Step 1: Let $X$ be the number of birds (out of $8$) with mass exceeding $0.25$ kg. Then $X \sim B(8, 0.3)$. Step 2: $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$. Step 3: $P(X = 0) = 0.7^8 = 0.05765$. Step 4: $P(X = 1) = \binom{8}{1} \cdot 0.3 \cdot 0.7^7 = 8 \cdot 0.3 \cdot 0.08235 = 0.19765$. Step 5: $P(X = 2) = \binom{8}{2} \cdot 0.3^2 \cdot 0.7^6 = 28 \cdot 0.09 \cdot 0.11765 = 0.29648$. Step 6: Sum $\approx 0.05765 + 0.19765 + 0.29648 = 0.55178 \approx 0.552$.
    Method:

    Use the binomial pmf for $X = 0, 1, 2$ and add them together.
    Examiner tips

    • Store intermediate values precisely; round only at the end
    • Note 'fewer than 3' means $X \leq 2$