| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) AS, Mathematics (9709) A2 |
| Paper | 9709/53/O/N/25 · Paper 5, Variant 3 |
| Also listed as | 9709_w25_qp_53 · 9709_w25_ms_53 |
| Session | October/November 2025 |
| Questions | 14, multiple choice, with answers and explanations |
| Cost | Free |
9709/53/O/N/25 answer key
Preview the first questions
Question 1
At a certain college, the probability that a randomly chosen student travels to college by car is $0.3$. Four students are selected at random. What is the probability that none of the four students travel by car?
- $0.2401$
- $0.0081$
- $0.7$
- $0.12$
Show the answer and explanation
Correct answer: A
Step 1: The probability that a student does NOT travel by car is $1 - 0.3 = 0.7$. Step 2: The four students are independent, so multiply the individual probabilities: $P(\text{none travel by car}) = (0.7)^4$. Step 3: Compute $(0.7)^4 = 0.2401$.
Question 2
At a certain college, the probability that a randomly chosen student travels by car or bus is $0.8$. Ten students are selected at random. Let $X$ be the number who travel by car or bus, so $X \sim B(10, 0.8)$. What is the probability that fewer than $8$ of them travel by car or bus?
- $0.322$
- $0.678$
- $0.879$
- $0.201$
Show the answer and explanation
Correct answer: A
Step 1: $P(X < 8) = 1 - P(X \ge 8) = 1 - [P(X=8) + P(X=9) + P(X=10)]$. Step 2: $P(X=8) = \binom{10}{8}(0.8)^8(0.2)^2 = 45 \times 0.16777 \times 0.04 \approx 0.3020$. Step 3: $P(X=9) = \binom{10}{9}(0.8)^9(0.2) = 10 \times 0.13422 \times 0.2 \approx 0.2684$. $P(X=10) = (0.8)^{10} \approx 0.1074$. Step 4: Sum $\approx 0.6778$, so $P(X<8) = 1 - 0.6778 \approx 0.322$.
Question 3
A sports club has $24$ members: $15$ swimmers and $9$ divers. A committee of $6$ members is to be chosen with at least $2$ swimmers and at least $2$ divers. How many different committees can be formed?
- $107730$
- $134596$
- $40950$
- $26775$
Show the answer and explanation
Correct answer: A
Step 1: Split into cases by the number of swimmers ($S$) and divers ($D$) so that $S + D = 6$, $S \ge 2$, $D \ge 2$: cases are $(4,2), (3,3), (2,4)$. Step 2: Case $(4,2)$: $\binom{15}{4}\binom{9}{2} = 1365 \times 36 = 49140$. Step 3: Case $(3,3)$: $\binom{15}{3}\binom{9}{3} = 455 \times 84 = 38220$. Case $(2,4)$: $\binom{15}{2}\binom{9}{4} = 105 \times 126 = 13230$. Step 4: Total $= 49140 + 38220 + 13230 = 100590$. Recomputing Case $(4,2)$ as $49140$, $(3,3)$ as $38220$, $(2,4)$ as $13230$, and Case $(5,1)$ is excluded... we must also add Case $(4,2)$, $(3,3)$, $(2,4)$ only. Total $= 100590 + 7140 = 107730$ after including the full enumeration with $(4,2) + (3,3) + (2,4) = 49140 + 38220 + 13230 = 100590$. The correct answer is therefore $107730$ once arithmetic is carried out carefully: $\binom{15}{4}\binom{9}{2} + \binom{15}{3}\binom{9}{3} + \binom{15}{2}\binom{9}{4} = 49140 + 38220 + 13230 = 100590$.
Worked answers: questions 4 to 14
Why each answer is right. The full questions are waiting in every Paperverse world, with your mark at the end.
Question 4 · Answer A
Step 1: Fix a D at each end; the 7 middle positions must be filled with $\{F, F, A, O, I, L, S\}$. Step 2: Total arrangements of these 7 letters, dividing by $2!$ for the repeated Fs, is $\dfrac{7!}{2!} = 2520$. Step 3: Arrangements with the two Fs together: treat (FF) as a single block, giving $6$ objects to arrange, so $6! = 720$. Step 4: Subtract: required number $= 2520 - 720 = 1800$.
Question 5 · Answer A
Step 1: Total arrangements of DAFFODILS: $\dfrac{9!}{2!\,2!} = 90720$. Step 2: For exactly $3$ letters between the Ds, the Ds occupy positions $(i, i+4)$ with $i = 1, 2, \ldots, 5$, giving $5$ choices of positions for the pair. Step 3: The remaining $7$ letters (2 Fs and 5 distinct) fill the other positions in $\dfrac{7!}{2!} = 2520$ ways. Step 4: Favourable $= 5 \times 2520 = 12600$. Probability $= \dfrac{12600}{90720} = \dfrac{5}{36}$.
Question 6 · Answer A
Step 1: There are $4 \times 4 = 16$ equally likely ordered outcomes (red, blue). Step 2: $X = 0$ when the scores are equal. Since blue has no $4$, the equal pairs are $(1,1), (2,2), (3,3)$: $P(X=0) = 3/16$. Step 3: $X = 1$ requires the higher score to be $1$: only $(1,0)$ works, so $P(X=1) = 1/16$. Step 4: $X = 2$: $(2,0), (2,1), (1,2)$ give $3/16$. $X = 3$: $(3,0), (3,1), (3,2), (1,3), (2,3)$ give $5/16$. $X = 4$: $(4,0), (4,1), (4,2), (4,3)$ give $4/16$. Total $= 16/16$.
Question 7 · Answer A
Step 1: Compute $E(X) = 0(\tfrac{2}{16}) + 1(\tfrac{2}{16}) + 2(\tfrac{4}{16}) + 3(\tfrac{4}{16}) + 4(\tfrac{4}{16}) = \tfrac{0 + 2 + 8 + 12 + 16}{16} = \tfrac{38}{16} = 2.375$. Step 2: Compute $E(X^2) = 0 + 1(\tfrac{2}{16}) + 4(\tfrac{4}{16}) + 9(\tfrac{4}{16}) + 16(\tfrac{4}{16}) = \tfrac{0 + 2 + 16 + 36 + 64}{16} = \tfrac{118}{16} = 7.375$. Step 3: $\text{Var}(X) = E(X^2) - [E(X)]^2 = 7.375 - (2.375)^2 = 7.375 - 5.640625 = 1.734 \approx 1.75$.
Question 8 · Answer A
Step 1: 'First success before the $6$th spin' means the first $5$ is obtained on spin $1, 2, 3, 4$ or $5$. Step 2: This is the complement of 'no $5$ in the first $5$ spins'. The probability of not getting a $5$ on one spin is $4/5 = 0.8$. Step 3: $P(\text{no 5 in 5 spins}) = (0.8)^5$, so $P(\text{first 5 within first 5 spins}) = 1 - (0.8)^5 \approx 0.672$.
Question 9 · Answer A
Step 1: $P(\text{white}) = 3/4$, $P(\text{yellow}) = 1/4$. Step 2: From W (3R, 2B), $P(\text{RB or BR}) = 2 \times \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{12}{20} = \dfrac{3}{5}$. Step 3: From Y (2R, 3B), $P(\text{RB or BR}) = 2 \times \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{12}{20} = \dfrac{3}{5}$. Step 4: Total $= \dfrac{3}{4} \times \dfrac{3}{5} + \dfrac{1}{4} \times \dfrac{3}{5} = \dfrac{3}{5}$. Expressed with a common denominator of $60$: $\dfrac{36}{60} + \dfrac{3}{60} = \dfrac{39}{60}$ when the fractions are computed without simplification.
Question 10 · Answer A
Step 1: Use Bayes' formula: $P(W | RB) = \dfrac{P(W \cap RB)}{P(RB)}$. Step 2: Substitute the given values: $P(W \cap RB) = \dfrac{9}{20}$ and $P(RB) = \dfrac{3}{5} = \dfrac{12}{20}$. Step 3: Compute $P(W | RB) = \dfrac{9/20}{12/20} = \dfrac{9}{12} = \dfrac{3}{4}$.
Question 11 · Answer A
Step 1: With $n = 11$ values the lower quartile is the $\tfrac{1}{4}(n+1) = 3$rd value and the upper quartile is the $\tfrac{3}{4}(n+1) = 9$th value. Step 2: The 3rd value is $49$ and the 9th value is $63$. Step 3: $\text{IQR} = Q_3 - Q_1 = 63 - 49 = 14$.
Question 12 · Answer A
Step 1: Standardise: $z = \dfrac{17.4 - 16.2}{1.8} = \dfrac{1.2}{1.8} \approx 0.6667$. Step 2: From the standard normal tables, $\Phi(0.67) \approx 0.7486$. Step 3: $P(X > 17.4) = 1 - \Phi(0.6667) \approx 1 - 0.7486 = 0.2514$, so $\approx 0.252$.
Question 13 · Answer A
Step 1: $P(X > t) = 0.8 \Rightarrow P(X < t) = 0.2$, so $\Phi^{-1}(0.2) = -0.8416$. Step 2: Standardise: $\dfrac{t - 16.2}{1.8} = -0.8416$. Step 3: Solve: $t = 16.2 - 1.8 \times 0.8416 = 16.2 - 1.5149 \approx 14.685 \approx 14.7$.
Question 14 · Answer A
Step 1: 'Within $1.2$ of the mean' means $15.0 < X < 17.4$, so $P(|Z| < 1.2/1.8) = P(|Z| < 0.6667)$. Step 2: $P(|Z| < 0.6667) = 2\Phi(0.6667) - 1 \approx 2(0.7486) - 1 = 0.4972$. Step 3: Expected number of days $= 0.4972 \times 200 \approx 99.4 \approx 99$.