| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric and binomial distributions with standard probability calculations. Part (a) uses the geometric distribution formula directly, (b) requires summing a short geometric series or using the complement, and (c) combines basic dice probability with binomial distribution. All parts are routine textbook exercises requiring recall of formulas rather than problem-solving insight, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\left(\frac{5}{6}\right)^7 \times \frac{1}{6} = \right] 0.0465, \frac{78125}{1679616}\) | B1 | \(0.0465 \leqslant p < 0.04652\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 6) = 1 - \left(\frac{5}{6}\right)^5\) or \(\frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)\) | M1 | \(1 - p^n\), \(0 < p < 1\), \(n = 4, 5, 6\) or sum of 4, 5 or 6 terms \(p \times (1-p)^n\) for \(n = 0,1,2,3,4(5)\) |
| \(0.598, \frac{4651}{7776}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| [Probability of total less than 4 is] \(\frac{3}{36}\) or \(\frac{1}{12}\) | B1 | SOI |
| \([1 - P(0, 1, 2)]\) \(= 1 - \left({}^{10}C_0\left(\frac{1}{12}\right)^0\left(\frac{11}{12}\right)^{10} + {}^{10}C_1\left(\frac{1}{12}\right)^1\left(\frac{11}{12}\right)^9 + {}^{10}C_2\left(\frac{1}{12}\right)^2\left(\frac{11}{12}\right)^8\right)\) | M1 | One term \({}^{10}C_x \; p^x(1-p)^{10-x}\), for \(0 < x < 10\), \(0 < p < 1\) |
| \(1 - (0.418904 + 0.380822 + 0.155791)\) | A1 FT | Correct expression. Accept unsimplified. |
| \(0.0445\) | A1 | \(0.04448 \leqslant p < 0.0445\) |
## Question 4:
**Part 4(a):**
$\left[\left(\frac{5}{6}\right)^7 \times \frac{1}{6} = \right] 0.0465, \frac{78125}{1679616}$ | B1 | $0.0465 \leqslant p < 0.04652$
**Part 4(b):**
$P(X < 6) = 1 - \left(\frac{5}{6}\right)^5$ or $\frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)$ | M1 | $1 - p^n$, $0 < p < 1$, $n = 4, 5, 6$ or sum of 4, 5 or 6 terms $p \times (1-p)^n$ for $n = 0,1,2,3,4(5)$
$0.598, \frac{4651}{7776}$ | A1 |
**Part 4(c):**
[Probability of total less than 4 is] $\frac{3}{36}$ or $\frac{1}{12}$ | B1 | SOI
$[1 - P(0, 1, 2)]$ $= 1 - \left({}^{10}C_0\left(\frac{1}{12}\right)^0\left(\frac{11}{12}\right)^{10} + {}^{10}C_1\left(\frac{1}{12}\right)^1\left(\frac{11}{12}\right)^9 + {}^{10}C_2\left(\frac{1}{12}\right)^2\left(\frac{11}{12}\right)^8\right)$ | M1 | One term ${}^{10}C_x \; p^x(1-p)^{10-x}$, for $0 < x < 10$, $0 < p < 1$
$1 - (0.418904 + 0.380822 + 0.155791)$ | A1 FT | Correct expression. Accept unsimplified.
$0.0445$ | A1 | $0.04448 \leqslant p < 0.0445$
---4 Ramesh throws an ordinary fair 6-sided die.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that he obtains a 4 for the first time on his 8th throw.
\item Find the probability that it takes no more than 5 throws for Ramesh to obtain a 4 .\\
Ramesh now repeatedly throws two ordinary fair 6-sided dice at the same time. Each time he adds the two numbers that he obtains.
\item For 10 randomly chosen throws of the two dice, find the probability that Ramesh obtains a total of less than 4 on at least three throws.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q4 [7]}}