| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Second success on trial n |
| Difficulty | Moderate -0.8 Part (a) requires a straightforward geometric distribution calculation using the complement rule or summing probabilities. Part (b) is a standard 'second success on trial n' problem requiring recognition that the first success occurs in the first 7 trials and the second on trial 8, which is a direct application of the negative binomial formula. Both parts are routine applications of standard formulas with no problem-solving insight required. |
| Spec | 5.02f Geometric distribution: conditions5.02g 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 |
|---|---|---|
| \([P(X < 8) =] \ 1 - \left(\dfrac{5}{6}\right)^7\) | M1 | \(1 - b^d\), \(b = \dfrac{5}{6}, \dfrac{1}{6}\), \(d = 7, 8\); or \(1 - c^e - (1-c) \times c^{e-1}\), \(c = \dfrac{5}{6}, \dfrac{1}{6}\), \(e = 8, 9\) |
| \(= 0.721\) | A1 | \(0.720918\ldots \ \dfrac{201811}{279936}\); If M0 scored, SC B1 for \(0.7209\ldots\) or \(\dfrac{201811}{279936}\) only |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X < 8) =] \ \dfrac{1}{6} + \left(\dfrac{5}{6}\right)\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^2\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^3\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^5\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^6\left(\dfrac{1}{6}\right)\) | M1 | \(a + ba + b^2a + b^3a + b^4a + b^5a + b^6a \ [+b^7a]\); \(a = \dfrac{1}{6}\), \(\dfrac{5}{6}\), \(a + b = 1\) |
| \(= 0.721\) | A1 | \(0.720918\ldots \ \dfrac{201811}{279936}\); If M0 scored, SC B1 for \(0.7209\ldots\) or \(\dfrac{201811}{279936}\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{5}{6}\right)^6 \times \left(\frac{1}{6}\right)^2 \times 7\) | M1 | \(\left(\frac{5}{6}\right)^6 \times \left(\frac{1}{6}\right)^2 \times d\), where \(d\) is an integer \(\geq 1\), no inappropriate addition |
| \(0.0651\) | A1 | \(0.0651 \leqslant p < 0.06512\) |
## Question 1(a):
**Method 1:**
$[P(X < 8) =] \ 1 - \left(\dfrac{5}{6}\right)^7$ | M1 | $1 - b^d$, $b = \dfrac{5}{6}, \dfrac{1}{6}$, $d = 7, 8$; or $1 - c^e - (1-c) \times c^{e-1}$, $c = \dfrac{5}{6}, \dfrac{1}{6}$, $e = 8, 9$
$= 0.721$ | A1 | $0.720918\ldots \ \dfrac{201811}{279936}$; If M0 scored, **SC B1** for $0.7209\ldots$ or $\dfrac{201811}{279936}$ only
---
**Method 2:**
$[P(X < 8) =] \ \dfrac{1}{6} + \left(\dfrac{5}{6}\right)\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^2\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^3\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^5\left(\dfrac{1}{6}\right) + \left(\dfrac{5}{6}\right)^6\left(\dfrac{1}{6}\right)$ | M1 | $a + ba + b^2a + b^3a + b^4a + b^5a + b^6a \ [+b^7a]$; $a = \dfrac{1}{6}$, $\dfrac{5}{6}$, $a + b = 1$
$= 0.721$ | A1 | $0.720918\ldots \ \dfrac{201811}{279936}$; If M0 scored, **SC B1** for $0.7209\ldots$ or $\dfrac{201811}{279936}$ only
**Total: 2 marks**
# Question 1:
## Part 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{5}{6}\right)^6 \times \left(\frac{1}{6}\right)^2 \times 7$ | M1 | $\left(\frac{5}{6}\right)^6 \times \left(\frac{1}{6}\right)^2 \times d$, where $d$ is an integer $\geq 1$, no inappropriate addition |
| $0.0651$ | A1 | $0.0651 \leqslant p < 0.06512$ |
---1 Nicola throws an ordinary fair six-sided dice. The random variable $X$ is the number of throws that she takes to obtain a 6.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X < 8 )$.
\item Find the probability that Nicola obtains a 6 for the second time on her 8th throw.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad3a6a8a-23fe-415a-b2f4-7c49136ccc6c-02_2717_35_109_2012}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q1 [4]}}