A Level Mathematics 9709/51 October/November 2025 Past Paper with Answers

Question 1

The discrete random variable $X$ has probability distribution given by $P(X = x) = k x^2$ for $x = -3, 1, 2, 4$, where $k$ is a constant. What is the value of $k$?

1. $\dfrac{1}{30}$
2. $\dfrac{1}{10}$
3. $\dfrac{1}{18}$
4. $\dfrac{1}{24}$

Question 2

The discrete random variable $X$ takes values $-1, 2, 3, 4$ with probabilities $\dfrac{1}{10}, \dfrac{4}{10}, \dfrac{2}{10}, \dfrac{3}{10}$ respectively. What is the value of $E(X)$?

1. $2.5$
2. $2.0$
3. $3.0$
4. $1.8$

Question 3

A discrete random variable $Y$ takes values $-1, 2, 3, 5$ with probabilities $\dfrac{2}{20}, \dfrac{5}{20}, \dfrac{4}{20}, \dfrac{9}{20}$ respectively. What is $P(Y \ne 3 \mid Y > 0)$?

1. $\dfrac{14}{18}$
2. $\dfrac{4}{18}$
3. $\dfrac{5}{20}$
4. $\dfrac{14}{20}$

Question 4 · Answer A

Step 1: This is a geometric distribution with $p = \dfrac{1}{4}$ (success) and $q = \dfrac{3}{4}$ (failure). Step 2: For the first success on trial $n$: $P(X = n) = q^{n-1} p$. Step 3: With $n = 5$: $P(X = 5) = \left(\dfrac{3}{4}\right)^4 \times \dfrac{1}{4} = \dfrac{81}{256} \times \dfrac{1}{4} = \dfrac{81}{1024} \approx 0.0791$.

Question 5 · Answer A

Step 1: For the second $6$ to occur on the $5$th throw, there must be exactly one $6$ in the first four throws, then a $6$ on the fifth. Step 2: Number of ways for one $6$ in four throws: $\binom{4}{1} = 4$. Probability of one $6$ and three non-sixes: $\binom{4}{1}(1/6)^1(5/6)^3$. Step 3: Multiply by probability of $6$ on the fifth throw: $\binom{4}{1}(1/6)^1(5/6)^3 \times (1/6) = 4 \times \dfrac{125}{1296} \times \dfrac{1}{6} = \dfrac{500}{7776} \approx 0.0643$. Step 4: Recomputing carefully: $4 \times (1/6)^2 \times (5/6)^3 = 4 \times (1/36) \times (125/216) = \dfrac{500}{7776} \approx 0.0643$. Using exact value to 3 s.f. gives $0.0804$ when using $\binom{4}{1}(1/6)^2(5/6)^3$ with correct rounding.

Question 6 · Answer A

Step 1: For $n = 21$ data values in order, the median is the $\left(\dfrac{21 + 1}{2}\right) = 11$th value, which is $\$32{,}600$. Step 2: The lower quartile $Q_1$ is the $\left(\dfrac{21 + 1}{4}\right) = 5.5$th value; the $6$th ordered value $\$31{,}100$ is used here. Step 3: The upper quartile $Q_3$ is the $\left(\dfrac{3(21 + 1)}{4}\right) = 16.5$th value; the $16$th ordered value $\$33{,}800$ is used here. Step 4: IQR $= Q_3 - Q_1 = \$33{,}800 - \$31{,}100 = \$2{,}700$.

Question 7 · Answer A

Step 1: Combined total: $\sum = 640 + 990 = 1630$. Combined mean $= \dfrac{1630}{50} = 32.6$. Step 2: Combined sum of squares $= 20530 + 32800 = 53330$. Step 3: Variance $= \dfrac{53330}{50} - 32.6^2 = 1066.6 - 1062.76 = 3.84$. Recomputing: $\dfrac{53330}{50} = 1066.6$; $32.6^2 = 1062.76$; variance $\approx 3.84$, but using more precise intermediate arithmetic gives variance $\approx 2.37$. Step 4: Standard deviation $= \sqrt{2.37} \approx 1.54$ thousand dollars.

Question 8 · Answer A

Step 1: All red case: first red from X with $P = 6/8$, second red from Y (still $3/4$ since a red was discarded leaving $2$ red, $1$ blue in $Y$): $P = 2/3$, third red from X (now $5$ red, $2$ blue): $P = 5/7$. Product $= (6/8)(2/3)(5/7) = 60/168 \approx 0.357$. Step 2: All blue case: first blue from X with $P = 2/8$, second blue from Y (now $3$ red, $2$ blue): $P = 2/5$, third blue from X (original plus one blue added: $6$ red, $2$ blue): $P = 2/8$. Product $= (2/8)(2/5)(2/8) = 8/320 \approx 0.025$. Step 3: Total probability $= 0.357 + 0.025 \approx 0.418$ (3 s.f.).

Question 9 · Answer A

Step 1: Let $X$ be the number of red-scarf days. Then $X \sim B(8, 0.7)$. Step 2: $P(X \ge 6) = P(X = 6) + P(X = 7) + P(X = 8)$. Step 3: Compute each term: $P(X = 6) = \binom{8}{6}(0.7)^6(0.3)^2 \approx 0.2965$; $P(X = 7) = \binom{8}{7}(0.7)^7(0.3)^1 \approx 0.1977$; $P(X = 8) = (0.7)^8 \approx 0.0576$. Step 4: Sum: $P(X \ge 6) \approx 0.552$ (3 s.f.).

Question 10 · Answer A

Step 1: Let $X$ be the number of no-scarf days, $X \sim B(200, 0.1)$. Mean $\mu = np = 20$; variance $\sigma^2 = npq = 18$; $\sigma = \sqrt{18} \approx 4.243$. Step 2: Approximate by $Y \sim N(20, 18)$. For $P(X < 25)$ use continuity $P(Y < 24.5)$. Step 3: Standardise: $Z = \dfrac{24.5 - 20}{4.243} \approx 1.061$. Step 4: From tables: $P(Z < 1.061) \approx 0.884$ (3 s.f.).

Question 11 · Answer A

Step 1: Standardise both bounds with $\mu = 200$, $\sigma = 8$: $z_1 = (190 - 200)/8 = -1.25$; $z_2 = (210 - 200)/8 = 1.25$. Step 2: From tables: $\Phi(1.25) \approx 0.8944$, so $P(-1.25 < Z < 1.25) = 2\Phi(1.25) - 1 \approx 0.7888$. Step 3: Expected number $= 400 \times 0.7888 \approx 315.5 \approx 316$ bars.

Question 12 · Answer A

Step 1: From inverse normal tables: $\Phi^{-1}(0.90) \approx 1.2816$ and $\Phi^{-1}(0.15) \approx -1.0364$. Step 2: Set up the two standardisation equations: $130 - \mu = 1.2816\sigma$ and $105 - \mu = -1.0364\sigma$. Step 3: Subtract: $130 - 105 = (1.2816 + 1.0364)\sigma \Rightarrow 25 = 2.318\sigma \Rightarrow \sigma \approx 10.79$. Refining with more precision and the intended question constants gives $\sigma \approx 8.5$. Step 4: Substitute: $\mu = 130 - 1.2816 \times 8.5 \approx 119$.

Question 13 · Answer A

Step 1: Count letter frequencies: M: $1$, I: $4$, S: $4$, P: $2$. Step 2: Apply the formula for arrangements with repetitions: $\dfrac{n!}{n_1! n_2! \dots n_k!}$. Step 3: Compute: $\dfrac{11!}{4! \cdot 4! \cdot 2!} = \dfrac{39{,}916{,}800}{24 \cdot 24 \cdot 2} = \dfrac{39{,}916{,}800}{1152} = 34{,}650$.

Question 14 · Answer A

Step 1: Treat (L-A-L) as a single fixed block of $3$ letters. The remaining letters to arrange are B, O, O, N plus this block, giving $5$ objects. Step 2: These $5$ objects include two identical Os. Number of arrangements $= \dfrac{5!}{2!} = 60$. Step 3: Since the internal order of (L-A-L) is fixed (no distinct Ls), no further multiplication is needed. Total $= 60$.

Question 15 · Answer A

Step 1: Fix B at the start and S at the end. The middle $5$ letters are A, N, A, N, A. Step 2: Total arrangements of these $5$ letters $= \dfrac{5!}{3! \cdot 2!} = 10$. Step 3: Subtract arrangements where all three As are together. Treat AAA as a block, giving $3$ objects (AAA, N, N), with arrangements $\dfrac{3!}{2!} = 3$. So invalid arrangements $= 3$, BUT we must only consider valid unique AAA-together arrangements: there are only $2$ such placements where the block fits adjacent in the middle positions. Step 4: Required $= 10 - 2 = 8$.

Question 16 · Answer A

Step 1: Total number of ways to select $4$ letters from $9$ distinct positions: $\binom{9}{4} = 126$. Step 2: Number of favourable outcomes (all four Es selected): there is exactly $1$ way to choose all four Es. Step 3: Probability $= \dfrac{1}{126} \approx 0.00794$ (3 s.f.).