| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| 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 application of geometric and binomial distributions with clear probability models. Part (a) uses geometric distribution with standard 'before trial n' calculation, part (b) is binomial complement, and part (c) requires multinomial/permutation counting. All are routine S1-level questions requiring formula application rather than problem-solving insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \([P(X < 6) = P(X \leqslant 5) =] \ 1 - 0.8^5\) | M1 | \(1 - 0.8^r\), \(r = 5, 6\) |
| \(= 0.672\) | A1 | |
| Method 2: \([P(X < 6) = P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) =]\) \(\frac{1}{5}+\frac{4}{5}\times\frac{1}{5}+\left(\frac{4}{5}\right)^2\times\frac{1}{5}+\left(\frac{4}{5}\right)^3\times\frac{1}{5}+\left(\frac{4}{5}\right)^4\times\frac{1}{5}\) | M1 | Condone an extra term \(\left(\frac{4}{5}\right)^5\times\frac{1}{5}\). First, last and one of the 3 middle terms implies M1 |
| \(= 0.672\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \([1 - P(0,1,2)]\) \(= 1 - \left({}^{12}C_0(0.8)^{12} + {}^{12}C_1(0.2)(0.8)^{11} + {}^{12}C_2(0.2)^2(0.8)^{10}\right)\) | M1 | One term \({}^{12}C_x\,(p)^x(1-p)^{12-x}\), \(0 < p < 1\), \(x \neq 0,1,2\) |
| \([= 1-(0.06872+0.20615+0.28347)]\) | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer. Correct unsimplified expression or better |
| \(= 0.442\) | B1 | \(0.411 < p \leqslant 0.442\) WWW |
| Method 2: \([P(3,4,5,6,7,8,9,10,11,12) =]\) \({}^{12}C_3(0.2)^3(0.8)^9 + {}^{12}C_4(0.2)^4(0.8)^8 + \ldots + {}^{12}C_{11}(0.2)^{11}(0.8)^1 + {}^{12}C_{12}(0.2)^{12}\) | M1 | One term \({}^{12}C_x\,(p)^x(1-p)^{12-x}\), \(0 < p < 1\), \(x \neq 0,1,2\) |
| \([= 0.23622 + 0.13288 + \ldots + 1.966\times10^{-7} + 4.096\times10^{-9}]\) | A1 | Correct expression, accept unsimplified, leading to final answer. Accept first, last and 8 of the middle terms |
| \(= 0.442\) | B1 | \(0.411 < p \leqslant 0.442\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.2)^5 \times 5!\) | M1 | \((0.2)^5 \times s\), \(s\) a positive integer. 1 may be implied |
| M1 | \(t \times 5!\) where \(0 < t < 1\) | |
| \(= 0.0384,\ \frac{24}{625}\) | A1 | |
| Alternative Method: \(\dfrac{{}^5C_1 \times {}^4C_1 \times {}^3C_1 \times {}^2C_1 \times [{}^1C_1]}{({}^5C_1)^5}\) | M1 | \(({}^5C_1)^5\) or \(5^5\) as denominator |
| M1 | \({}^5C_1 \times {}^4C_1 \times {}^3C_1 \times {}^2C_1 \times [{}^1C_1]\) or \(5!\) as numerator | |
| \(= 0.0384,\ \frac{24}{625}\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $[P(X < 6) = P(X \leqslant 5) =] \ 1 - 0.8^5$ | M1 | $1 - 0.8^r$, $r = 5, 6$ |
| $= 0.672$ | A1 | |
| **Method 2:** $[P(X < 6) = P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) =]$ $\frac{1}{5}+\frac{4}{5}\times\frac{1}{5}+\left(\frac{4}{5}\right)^2\times\frac{1}{5}+\left(\frac{4}{5}\right)^3\times\frac{1}{5}+\left(\frac{4}{5}\right)^4\times\frac{1}{5}$ | M1 | Condone an extra term $\left(\frac{4}{5}\right)^5\times\frac{1}{5}$. First, last and one of the 3 middle terms implies M1 |
| $= 0.672$ | A1 | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $[1 - P(0,1,2)]$ $= 1 - \left({}^{12}C_0(0.8)^{12} + {}^{12}C_1(0.2)(0.8)^{11} + {}^{12}C_2(0.2)^2(0.8)^{10}\right)$ | M1 | One term ${}^{12}C_x\,(p)^x(1-p)^{12-x}$, $0 < p < 1$, $x \neq 0,1,2$ |
| $[= 1-(0.06872+0.20615+0.28347)]$ | A1 | Correct expression, accept unsimplified, no terms omitted, leading to final answer. Correct unsimplified expression or better |
| $= 0.442$ | B1 | $0.411 < p \leqslant 0.442$ WWW |
| **Method 2:** $[P(3,4,5,6,7,8,9,10,11,12) =]$ ${}^{12}C_3(0.2)^3(0.8)^9 + {}^{12}C_4(0.2)^4(0.8)^8 + \ldots + {}^{12}C_{11}(0.2)^{11}(0.8)^1 + {}^{12}C_{12}(0.2)^{12}$ | M1 | One term ${}^{12}C_x\,(p)^x(1-p)^{12-x}$, $0 < p < 1$, $x \neq 0,1,2$ |
| $[= 0.23622 + 0.13288 + \ldots + 1.966\times10^{-7} + 4.096\times10^{-9}]$ | A1 | Correct expression, accept unsimplified, leading to final answer. Accept first, last and 8 of the middle terms |
| $= 0.442$ | B1 | $0.411 < p \leqslant 0.442$ |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.2)^5 \times 5!$ | M1 | $(0.2)^5 \times s$, $s$ a positive integer. 1 may be implied |
| | M1 | $t \times 5!$ where $0 < t < 1$ |
| $= 0.0384,\ \frac{24}{625}$ | A1 | |
| **Alternative Method:** $\dfrac{{}^5C_1 \times {}^4C_1 \times {}^3C_1 \times {}^2C_1 \times [{}^1C_1]}{({}^5C_1)^5}$ | M1 | $({}^5C_1)^5$ or $5^5$ as denominator |
| | M1 | ${}^5C_1 \times {}^4C_1 \times {}^3C_1 \times {}^2C_1 \times [{}^1C_1]$ or $5!$ as numerator |
| $= 0.0384,\ \frac{24}{625}$ | A1 | |7 A children's wildlife magazine is published every Monday. For the next 12 weeks it will include a model animal as a free gift. There are five different models: tiger, leopard, rhinoceros, elephant and buffalo, each with the same probability of being included in the magazine.
Sahim buys one copy of the magazine every Monday.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the first time that the free gift is an elephant is before the 6th Monday.
\item Find the probability that Sahim will get more than two leopards in the 12 magazines.
\item Find the probability that after 5 weeks Sahim has exactly one of each animal.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q7 [8]}}