| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | March |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution formula with clearly stated parameters (p=1/5). Part (a) requires direct substitution into (1-p)^(n-1)×p, and part (b) requires either summing a short geometric series or using 1-(1-p)^n. Both are standard textbook exercises with no problem-solving or conceptual challenges beyond recognizing the distribution type. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[\left(\frac{4}{5}\right)^7 \frac{1}{5} =\right] \frac{16384}{390625}\) or \(0.0419[43\ldots]\) | B1 | Evaluated, final answer. |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1-\left(\frac{4}{5}\right)^5\) or \(\frac{1}{5}+\frac{4}{5}\times\frac{1}{5}\left(\frac{4}{5}\right)^2\times\frac{1}{5}+\left(\frac{4}{5}\right)^3\times\frac{1}{5}+\left(\frac{4}{5}\right)^4\times\frac{1}{5}\) | M1 | \(1-p^n\), \(n=5,6\); or \(p+pq+pq^2+pq^3+pq^4\) \((+pq^5)\); \(0 |
| \(\frac{2101}{3125}\) or \(0.672[32]\) | A1 | Final answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{1}{5}\right)^5 + {}^5C_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right) + {}^5C_3\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^2 + {}^5C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^3 + {}^5C_1\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^4\) | M1 | \((p)^5+{}^5C_4(p)^4(q)+{}^5C_3(p)^3(q)^2+{}^5C_2(p)^2(q)^3+{}^5C_1(p)(q)^4\); or \((p)^6+{}^6C_5(p)^5(q)+{}^6C_4(p)^4(q)^2+{}^6C_3(p)^3(q)^3+{}^6C_2(p)^2(q)^4+{}^6C_1(p)(q)^5\); \(0 |
| \(\frac{2101}{3125}\) or \(0.672[32]\) | A1 | Final answer. |
| 2 |
## Question 1:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\left(\frac{4}{5}\right)^7 \frac{1}{5} =\right] \frac{16384}{390625}$ or $0.0419[43\ldots]$ | **B1** | Evaluated, final answer. |
| | **1** | |
### Part (b):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1-\left(\frac{4}{5}\right)^5$ or $\frac{1}{5}+\frac{4}{5}\times\frac{1}{5}\left(\frac{4}{5}\right)^2\times\frac{1}{5}+\left(\frac{4}{5}\right)^3\times\frac{1}{5}+\left(\frac{4}{5}\right)^4\times\frac{1}{5}$ | **M1** | $1-p^n$, $n=5,6$; or $p+pq+pq^2+pq^3+pq^4$ $(+pq^5)$; $0<p<1$, $p+q=1$. Sum of a geometric series may be used. |
| $\frac{2101}{3125}$ or $0.672[32]$ | **A1** | Final answer. |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{1}{5}\right)^5 + {}^5C_4\left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right) + {}^5C_3\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^2 + {}^5C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^3 + {}^5C_1\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^4$ | **M1** | $(p)^5+{}^5C_4(p)^4(q)+{}^5C_3(p)^3(q)^2+{}^5C_2(p)^2(q)^3+{}^5C_1(p)(q)^4$; or $(p)^6+{}^6C_5(p)^5(q)+{}^6C_4(p)^4(q)^2+{}^6C_3(p)^3(q)^3+{}^6C_2(p)^2(q)^4+{}^6C_1(p)(q)^5$; $0<p<1$, $p+q=1$. At least first, last and one intermediate term required to show pattern if not all terms stated. |
| $\frac{2101}{3125}$ or $0.672[32]$ | **A1** | Final answer. |
| | **2** | |1 A fair spinner with 5 sides numbered 1,2,3,4,5 is spun repeatedly. The score on each spin is the number on the side on which the spinner lands.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a score of 3 is obtained for the first time on the 8th spin.
\item Find the probability that fewer than 6 spins are required to obtain a score of 3 for the first time.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q1 [3]}}