| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success before/after trial n |
| Difficulty | Standard +0.3 This is a straightforward geometric distribution application requiring students to first calculate P(score ≥ 17) for three dice, then apply standard geometric probability formulas for 'first success on trial n' and 'success before trial n'. The dice probability calculation is routine enumeration, and the geometric distribution formulas are direct recall with simple substitution. |
| Spec | 5.02g 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 | Mark | Guidance |
| \([P(17 \text{ or } 18) =] \frac{4}{216} = \frac{1}{54}\), \(0.0185(185...)\) | B1 | May be seen used in calculation |
| \(P(X=6) = \left(\frac{53}{54}\right)^5 \cdot \frac{1}{54}\) | M1 | \(p(1-p)^5\), \(0 < p < 1\) |
| \(0.0169\) | A1 | \(0.01686 < p \leqslant 0.0169\); if A0 scored SC B1 for \(0.01686 < p \leqslant 0.0169\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X<8) =] 1 - \left(\frac{53}{54}\right)^7\) | M1 | \(1 - \left(\textit{their}\;\frac{53}{54} \text{ or } 0.98148\right)^r\) or correct, \(r = 7, 8\); \(0 < \textit{their}\; p < 1\) |
| \(0.123\) | A1 | \(0.1225 \leqslant p \leqslant 0.123\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{1}{54}\right)+\left(\frac{53}{54}\right)\left(\frac{1}{54}\right)+\left(\frac{53}{54}\right)^2\left(\frac{1}{54}\right)+\cdots+\left(\frac{53}{54}\right)^6\left(\frac{1}{54}\right)\) | M1 | \(q + pq + p^2q + p^3q + p^4q + p^5q[+p^6q]\), \(p+q=1\), \(0 |
| \(0.123\) | A1 | \(0.1225 \leqslant p \leqslant 0.123\) |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(17 \text{ or } 18) =] \frac{4}{216} = \frac{1}{54}$, $0.0185(185...)$ | B1 | May be seen used in calculation |
| $P(X=6) = \left(\frac{53}{54}\right)^5 \cdot \frac{1}{54}$ | M1 | $p(1-p)^5$, $0 < p < 1$ |
| $0.0169$ | A1 | $0.01686 < p \leqslant 0.0169$; if A0 scored **SC B1** for $0.01686 < p \leqslant 0.0169$ |
---
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X<8) =] 1 - \left(\frac{53}{54}\right)^7$ | M1 | $1 - \left(\textit{their}\;\frac{53}{54} \text{ or } 0.98148\right)^r$ or correct, $r = 7, 8$; $0 < \textit{their}\; p < 1$ |
| $0.123$ | A1 | $0.1225 \leqslant p \leqslant 0.123$ |
**Alternative method for 3(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{1}{54}\right)+\left(\frac{53}{54}\right)\left(\frac{1}{54}\right)+\left(\frac{53}{54}\right)^2\left(\frac{1}{54}\right)+\cdots+\left(\frac{53}{54}\right)^6\left(\frac{1}{54}\right)$ | M1 | $q + pq + p^2q + p^3q + p^4q + p^5q[+p^6q]$, $p+q=1$, $0<p,q<1$, $= \textit{their}\;\frac{53}{54}$ |
| $0.123$ | A1 | $0.1225 \leqslant p \leqslant 0.123$ |
---3 Three fair 6-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time repeatedly. The score on each throw is the sum of the numbers on the uppermost faces.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a score of 17 or more is first obtained on the 6th throw.
\item Find the probability that a score of 17 or more is obtained in fewer than 8 throws.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q3 [5]}}