| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | P(X ≤ n) or P(X < n) |
| Difficulty | Standard +0.3 This question tests standard geometric distribution calculations (parts a-b) and basic probability with binomial distribution (parts c-d). All parts use routine formulas with straightforward arithmetic—no novel problem-solving or proof required. The multi-part structure and conditional probability in part (c) add slight complexity, but these are textbook-level exercises for S1. |
| Spec | 2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=4) = (0.8)^3(0.2) = 0.1024,\ \frac{64}{625}\) | B1 | Condone 0.102 |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 6) = 1 - 0.8^5\) | M1 | \(1 - 0.8^d,\ d = 5, 6\) |
| \(= 0.672,\ \frac{2101}{3125}\) | A1 | 0.67232 to at least 3SF. If M0 awarded, SC B1 for \(\frac{2101}{3125}\) or 0.67232 only |
| Alternative Method: | ||
| \(P(X<6) = \frac{1}{5} + \frac{4}{5}\cdot\frac{1}{5} + \left(\frac{4}{5}\right)^2\frac{1}{5} + \left(\frac{4}{5}\right)^3\frac{1}{5} + \left(\frac{4}{5}\right)^4\frac{1}{5}\) | M1 | If answer correct, condone omission of 2 from 3 middle terms |
| \(= 0.672,\ \frac{2101}{3125}\) | A1 | 0.67232 to at least 3SF |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X > 0 \ | X \neq 2) = \frac{P(X > 0 \cap X \neq 2)}{P(X \neq 2)}\) | M1 |
| \(= \frac{14/25}{19/25}\) | M1 | \(P(X \neq 2) = \frac{19}{25}\), 0.76[0] seen as denominator of conditional probability fraction |
| \(= \frac{14}{19}\), 0.737 | A1 | Final answer \(= \frac{14}{19}\), 0.7368421… to at least 3SF. If A0, SC B1 for correct final answer www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table of outcomes shown, \(P(X > 0 \cap X \neq 2) = \frac{\text{Number of outcomes}(X > 0 \cap X \neq 2)}{\text{Number of outcomes } X \neq 2}\) | M1 | Number of outcomes \((X > 0 \cap X \neq 2) = \) 14 seen as numerator or denominator |
| M1 | Number of outcomes \((X \neq 2) = \) 19 seen as denominator | |
| \(\frac{14}{19}\), 0.737 | A1 | Final answer \(= \frac{14}{19}\), 0.7368421… to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(X > 2)] = 1 - P(0, 1, 2)\) with \(p = \frac{6}{25}\) | M1 | One term \({}^9C_x\,(p)^x(1-p)^{9-x}\), \(0 < p < 1\), \(0 < x < 9\) |
| \(1 - \left({}^9C_0\left(\frac{19}{25}\right)^9 + {}^9C_1\left(\frac{6}{25}\right)^1\left(\frac{19}{25}\right)^8 + {}^9C_2\left(\frac{6}{25}\right)^2\left(\frac{19}{25}\right)^7\right)\) \([1-(0.08459 + 0.2404 + 0.3037)]\) | A1 | \(1 - \left({}^9C_0(1-p)^9 + {}^9C_1(p)^1(1-p)^8 + {}^9C_2(p)^2(1-p)^7\right)\), \(0 < p < 1\). Correct expression from *their p*, no terms omitted |
| 0.371 | B1 | \(0.371 \leqslant p < 0.3715\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(X>2)] = P(3,4,5,6,7,8,9)\) with \(p = \frac{6}{25}\) | M1 | One term \({}^9C_x\,(p)^x(1-p)^{9-x}\) |
| \({}^9C_3\left(\frac{6}{25}\right)^3\left(\frac{19}{25}\right)^6 + {}^9C_4\left(\frac{6}{25}\right)^4\left(\frac{19}{25}\right)^5 + \ldots + {}^9C_8\left(\frac{6}{25}\right)^8\left(\frac{19}{25}\right)^1 + {}^9C_9\left(\frac{6}{25}\right)^9\) | A1 | Correct full expression from *their p* |
| 0.371 | B1 | \(0.371 \leqslant p < 0.3715\) |
## Question 4:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=4) = (0.8)^3(0.2) = 0.1024,\ \frac{64}{625}$ | B1 | Condone 0.102 |
| **Total** | **1** | |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 6) = 1 - 0.8^5$ | M1 | $1 - 0.8^d,\ d = 5, 6$ |
| $= 0.672,\ \frac{2101}{3125}$ | A1 | 0.67232 to at least 3SF. If M0 awarded, **SC B1** for $\frac{2101}{3125}$ or 0.67232 only |
| **Alternative Method:** | | |
| $P(X<6) = \frac{1}{5} + \frac{4}{5}\cdot\frac{1}{5} + \left(\frac{4}{5}\right)^2\frac{1}{5} + \left(\frac{4}{5}\right)^3\frac{1}{5} + \left(\frac{4}{5}\right)^4\frac{1}{5}$ | M1 | If answer correct, condone omission of 2 from 3 middle terms |
| $= 0.672,\ \frac{2101}{3125}$ | A1 | 0.67232 to at least 3SF |
| **Total** | **2** | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 0 \| X \neq 2) = \frac{P(X > 0 \cap X \neq 2)}{P(X \neq 2)}$ | M1 | $P(X > 0 \cap X \neq 2) = \frac{14}{25}$, 0.56[0] seen as numerator or denominator of conditional probability fraction |
| $= \frac{14/25}{19/25}$ | M1 | $P(X \neq 2) = \frac{19}{25}$, 0.76[0] seen as denominator of conditional probability fraction |
| $= \frac{14}{19}$, 0.737 | A1 | Final answer $= \frac{14}{19}$, 0.7368421… to at least 3SF. If A0, **SC B1** for correct final answer www |
**Alternative Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table of outcomes shown, $P(X > 0 \cap X \neq 2) = \frac{\text{Number of outcomes}(X > 0 \cap X \neq 2)}{\text{Number of outcomes } X \neq 2}$ | M1 | Number of outcomes $(X > 0 \cap X \neq 2) = $ 14 seen as numerator or denominator |
| | M1 | Number of outcomes $(X \neq 2) = $ 19 seen as denominator |
| $\frac{14}{19}$, 0.737 | A1 | Final answer $= \frac{14}{19}$, 0.7368421… to at least 3SF |
**Total: 3 marks**
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(X > 2)] = 1 - P(0, 1, 2)$ with $p = \frac{6}{25}$ | M1 | One term ${}^9C_x\,(p)^x(1-p)^{9-x}$, $0 < p < 1$, $0 < x < 9$ |
| $1 - \left({}^9C_0\left(\frac{19}{25}\right)^9 + {}^9C_1\left(\frac{6}{25}\right)^1\left(\frac{19}{25}\right)^8 + {}^9C_2\left(\frac{6}{25}\right)^2\left(\frac{19}{25}\right)^7\right)$ $[1-(0.08459 + 0.2404 + 0.3037)]$ | A1 | $1 - \left({}^9C_0(1-p)^9 + {}^9C_1(p)^1(1-p)^8 + {}^9C_2(p)^2(1-p)^7\right)$, $0 < p < 1$. Correct expression from *their p*, no terms omitted |
| 0.371 | B1 | $0.371 \leqslant p < 0.3715$ |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(X>2)] = P(3,4,5,6,7,8,9)$ with $p = \frac{6}{25}$ | M1 | One term ${}^9C_x\,(p)^x(1-p)^{9-x}$ |
| ${}^9C_3\left(\frac{6}{25}\right)^3\left(\frac{19}{25}\right)^6 + {}^9C_4\left(\frac{6}{25}\right)^4\left(\frac{19}{25}\right)^5 + \ldots + {}^9C_8\left(\frac{6}{25}\right)^8\left(\frac{19}{25}\right)^1 + {}^9C_9\left(\frac{6}{25}\right)^9$ | A1 | Correct full expression from *their p* |
| 0.371 | B1 | $0.371 \leqslant p < 0.3715$ |
**Total: 3 marks**
---4 A fair 5 -sided spinner has sides labelled 1, 2, 3, 4, 5. The spinner is spun repeatedly until a 2 is obtained on the side on which the spinner lands. The random variable $X$ denotes the number of spins required.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X = 4 )$.
\item Find $\mathrm { P } ( X < 6 )$.\\
Two fair 5 -sided spinners, each with sides labelled $1,2,3,4,5$, are spun at the same time. If the numbers obtained are equal, the score is 0 . Otherwise, the score is the higher number minus the lower number.
\item Find the probability that the score is greater than 0 given that the score is not equal to 2 .\\
The two spinners are spun at the same time repeatedly .
\item For 9 randomly chosen spins of the two spinners, find the probability that the score is greater than 2 on at least 3 occasions.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q4 [9]}}