| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Geometric then binomial separate scenarios |
| Difficulty | Moderate -0.8 Part (a) is a direct application of geometric distribution CDF requiring summing P(X<6) = 1-(5/6)^5, and part (b) is standard binomial probability P(X≥3) = 1-P(X≤2). Both are routine textbook exercises with straightforward probability calculations requiring only formula recall and basic arithmetic, making this easier than average. |
| 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 |
| \(1 - \left(\frac{5}{6}\right)^5\) or \(\frac{1}{6} + \frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^2 \times \frac{1}{6} + \left(\frac{5}{6}\right)^3 \times \frac{1}{6} + \left(\frac{5}{6}\right)^4 \times \frac{1}{6}\) | M1 | \(1 - p^n\), \(n = 5, 6\); or \(p + pq + pq^2 + pq^3 + pq^4\) \((+ pq^5)\); \(0 < p < 1\), \(p + q = 1\) |
| \(0.598\), \(\frac{4651}{7776}\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1 - P(0,1,2))\) \(1 - \left(\left(\frac{5}{6}\right)^{10} + {}^{10}C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^9 + {}^{10}C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^8\right)\) | M1 | \({}^{10}C_x \, p^x(1-p)^{10-x}\), \(0 < p < 1\), any \(p\), \(x \neq 0, 10\) |
| \(1 - (0.1615056 + 0.3230111 + 0.290710)\) | A1 | Correct expression, accept unsimplified, condone omission of final bracket |
| \(0.225\) | A1 | \(0.2247 < p \leq 0.225\), WWW |
| Total: 3 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \left(\frac{5}{6}\right)^5$ or $\frac{1}{6} + \frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^2 \times \frac{1}{6} + \left(\frac{5}{6}\right)^3 \times \frac{1}{6} + \left(\frac{5}{6}\right)^4 \times \frac{1}{6}$ | M1 | $1 - p^n$, $n = 5, 6$; or $p + pq + pq^2 + pq^3 + pq^4$ $(+ pq^5)$; $0 < p < 1$, $p + q = 1$ |
| $0.598$, $\frac{4651}{7776}$ | A1 | |
| | **Total: 2** | |
---
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1 - P(0,1,2))$ $1 - \left(\left(\frac{5}{6}\right)^{10} + {}^{10}C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^9 + {}^{10}C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^8\right)$ | M1 | ${}^{10}C_x \, p^x(1-p)^{10-x}$, $0 < p < 1$, any $p$, $x \neq 0, 10$ |
| $1 - (0.1615056 + 0.3230111 + 0.290710)$ | A1 | Correct expression, accept unsimplified, condone omission of final bracket |
| $0.225$ | A1 | $0.2247 < p \leq 0.225$, WWW |
| | **Total: 3** | |1 A fair six-sided die, with faces marked $1,2,3,4,5,6$, is thrown repeatedly until a 4 is obtained.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that obtaining a 4 requires fewer than 6 throws.\\
On another occasion, the die is thrown 10 times.
\item Find the probability that a 4 is obtained at least 3 times.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q1 [5]}}