| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Geometric then binomial separate scenarios |
| Difficulty | Moderate -0.3 This question tests standard geometric and binomial distribution applications with straightforward probability calculations. Part (a) requires finding P(score ≥ 8) = 15/36 then applying geometric distribution formula. Parts (b) and (c) are direct applications of geometric CDF and binomial probability respectively. While multi-part, each component is routine S1 material requiring no novel insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\dfrac{21}{36}\right)^4\left(\dfrac{15}{36}\right)\) | M1 | \((1-p)^4 \times p,\ 0 < p < 1\) |
| \(= \dfrac{12005}{248832},\ 0.0482\) | A1 | \(0.0482454\ldots\) to at least 3SF. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X \leq 4) =] 1 - \left(\frac{21}{36}\right)^4\) | M1 | \(1 - b^r\), \(b = \text{their } (1-p)\) in 2(a) or correct; \(r = 4, 5\) |
| \(= \frac{18335}{20736}\), \(0.884\) | A1 | \(0.884211\ldots\) to at least 3SF |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X \leq 4) =] \frac{15}{36} + \frac{15}{36} \times \frac{21}{36} + \frac{15}{36} \times \left(\frac{21}{36}\right)^2 + \frac{15}{36} \times \left(\frac{21}{36}\right)^3\) | M1 | \(p + p(1-p) + p(1-p)^2 + p(1-p)^3 \left[+ p(1-p)^4\right]\) FT from 2(a) or correct |
| \(= \frac{18335}{20736}\), \(0.884\) | A1 | \(0.884211\ldots\) to at least 3SF |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(0,1,2) =] \; {}^8C_0\left(\frac{5}{12}\right)^0\left(\frac{7}{12}\right)^8 + {}^8C_1\left(\frac{5}{12}\right)^1\left(\frac{7}{12}\right)^7 + {}^8C_2\left(\frac{5}{12}\right)^2\left(\frac{7}{12}\right)^6\) | M1 | One term \({}^8C_x(q)^x(1-q)^{8-x}\), \(0 < q < 1\), \(x \neq 0, 8\) |
| \(0.01341 + 0.07661 + 0.1915\) | A1 FT | Correct expression, accept unsimplified, no terms omitted leading to final answer. FT only with unsimplified expression. |
| \(= 0.282\) | B1 | \(0.2815 \leq q \leq 0.282\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([1 - P(3,4,5,6,7,8) =] 1 - \left({}^8C_3\left(\frac{5}{12}\right)^3\left(\frac{7}{12}\right)^5 + {}^8C_4\left(\frac{5}{12}\right)^4\left(\frac{7}{12}\right)^4 + \ldots + {}^8C_7\left(\frac{5}{12}\right)^7\left(\frac{7}{12}\right)^1 + {}^8C_8\left(\frac{5}{12}\right)^8\left(\frac{7}{12}\right)^0\right)\) | M1 | One term \({}^8C_x(q)^x(1-q)^{8-x}\), \(0 < q < 1\), \(x \neq 0, 8\) |
| A1 FT | Correct expression, accept unsimplified, no terms omitted. FT only with unsimplified expression. |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.282\) | B1 | \(0.2815 \leq q \leq 0.282\) |
| 3 |
**Question 2:**
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\dfrac{21}{36}\right)^4\left(\dfrac{15}{36}\right)$ | M1 | $(1-p)^4 \times p,\ 0 < p < 1$ |
| $= \dfrac{12005}{248832},\ 0.0482$ | A1 | $0.0482454\ldots$ to at least 3SF. |
| | **Total: 2** | |
## Question 2(b):
**Method 1:**
$[P(X \leq 4) =] 1 - \left(\frac{21}{36}\right)^4$ | M1 | $1 - b^r$, $b = \text{their } (1-p)$ in 2(a) or correct; $r = 4, 5$
$= \frac{18335}{20736}$, $0.884$ | A1 | $0.884211\ldots$ to at least 3SF
| **2** |
**Method 2:**
$[P(X \leq 4) =] \frac{15}{36} + \frac{15}{36} \times \frac{21}{36} + \frac{15}{36} \times \left(\frac{21}{36}\right)^2 + \frac{15}{36} \times \left(\frac{21}{36}\right)^3$ | M1 | $p + p(1-p) + p(1-p)^2 + p(1-p)^3 \left[+ p(1-p)^4\right]$ FT from 2(a) or correct
$= \frac{18335}{20736}$, $0.884$ | A1 | $0.884211\ldots$ to at least 3SF
| **2** |
---
## Question 2(c):
**Method 1:**
$[P(0,1,2) =] \; {}^8C_0\left(\frac{5}{12}\right)^0\left(\frac{7}{12}\right)^8 + {}^8C_1\left(\frac{5}{12}\right)^1\left(\frac{7}{12}\right)^7 + {}^8C_2\left(\frac{5}{12}\right)^2\left(\frac{7}{12}\right)^6$ | M1 | One term ${}^8C_x(q)^x(1-q)^{8-x}$, $0 < q < 1$, $x \neq 0, 8$
$0.01341 + 0.07661 + 0.1915$ | A1 FT | Correct expression, accept unsimplified, no terms omitted leading to final answer. FT only with unsimplified expression.
$= 0.282$ | B1 | $0.2815 \leq q \leq 0.282$
**Method 2:**
$[1 - P(3,4,5,6,7,8) =] 1 - \left({}^8C_3\left(\frac{5}{12}\right)^3\left(\frac{7}{12}\right)^5 + {}^8C_4\left(\frac{5}{12}\right)^4\left(\frac{7}{12}\right)^4 + \ldots + {}^8C_7\left(\frac{5}{12}\right)^7\left(\frac{7}{12}\right)^1 + {}^8C_8\left(\frac{5}{12}\right)^8\left(\frac{7}{12}\right)^0\right)$ | M1 | One term ${}^8C_x(q)^x(1-q)^{8-x}$, $0 < q < 1$, $x \neq 0, 8$
| A1 FT | Correct expression, accept unsimplified, no terms omitted. FT only with unsimplified expression.
$= 1 - (0.2736 + 0.2443 + \ldots + 0.01017 + 9.084 \times 10^{-4})$
$= 0.282$ | B1 | $0.2815 \leq q \leq 0.282$
| **3** |
---2 Hazeem repeatedly throws two ordinary fair 6-sided dice at the same time. On each occasion, the score is the sum of the two numbers that she obtains.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that it takes exactly 5 throws of the two dice for Hazeem to obtain a score of 8 or more.
\item Find the probability that it takes no more than 4 throws of the two dice for Hazeem to obtain a score of 8 or more.
\item For 8 randomly chosen throws of the two dice, find the probability that Hazeem obtains a score of 8 or more on fewer than 3 occasions.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q2 [7]}}