| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Combined probability with other distributions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question combining basic probability (conditional probability from a contingency table) with standard binomial distribution calculations and normal approximation. Part (a) requires simple table reading and division; part (b)(i) is direct binomial calculation with small n; part (b)(ii) is routine normal approximation to binomial. All techniques are standard S1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Quality of broadband service | |||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Excellent | Good | Poor | |
| \multirow{3}{*}{Village} | Reeta | 75 | 118 | 32 |
| \cline { 2 - 5 } | Shan | 223 | 177 | 40 |
| \cline { 2 - 5 } | Teber | 12 | 60 | 63 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{40}{800}\) or \(\frac{1}{20}\) or \(0.05\) | B1 | |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{177}{223+177+40}\) | M1 | Their \(223+177+40\) seen as denominator of fraction in the final answer, accept unsimplified |
| \(\frac{177}{440}\) or \(0.402\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(G \mid S) = \dfrac{P(G \cap S)}{P(S)} = \dfrac{\frac{177}{800}}{\frac{223+177+40}{800}} = \dfrac{\frac{177}{800}}{\frac{440}{800}} = \dfrac{\frac{177}{800}}{\frac{11}{20} \text{ or } 0.55}\) | M1 | Their \(P(S)\) seen as denominator of fraction in the final answer, accept unsimplified |
| \(\frac{177}{440}\) or \(0.402\) | A1 | CAO |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(0,1,2) = {^{10}C_0}(0.35)^0(0.65)^{10} + {^{10}C_1}(0.35)^1(0.65)^9 + {^{10}C_2}(0.35)^2(0.65)^8\) | M1 | One term: \(^{10}C_x p^x(1-p)^{10-x}\) for \(0 < x < 10\), any \(0 < p < 1\) |
| \(0.013463 + 0.072492 + 0.17565\) | A1 | Correct unsimplified expression, or better |
| \(0.262\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= 120 \times 0.35\ [= 42]\); Variance \(= 120 \times 0.35 \times 0.65\ [= 27.3]\) | B1 | Correct mean and variance seen, allow unsimplified |
| \(P(X > 32) = P\!\left(Z > \dfrac{32.5 - 42}{\sqrt{27.3}}\right) = P(Z > -1.818)\) | M1 | Substituting their mean and variance into \(\pm\)standardisation formula (any number), condone \(\sigma^2\) or \(\sqrt{\sigma}\) |
| M1 | Using continuity correction \(31.5\) or \(32.5\) | |
| \(\Phi(1.818)\) | M1 | Appropriate area \(\Phi\), from final process, must be probability |
| \(0.966\) | A1 | \(0.965 \leqslant p \leqslant 0.966\) |
| 5 |
## Question 7(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{40}{800}$ or $\frac{1}{20}$ or $0.05$ | B1 | |
| | **1** | |
---
## Question 7(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{177}{223+177+40}$ | M1 | Their $223+177+40$ seen as denominator of fraction in the final answer, accept unsimplified |
| $\frac{177}{440}$ or $0.402$ | A1 | CAO |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(G \mid S) = \dfrac{P(G \cap S)}{P(S)} = \dfrac{\frac{177}{800}}{\frac{223+177+40}{800}} = \dfrac{\frac{177}{800}}{\frac{440}{800}} = \dfrac{\frac{177}{800}}{\frac{11}{20} \text{ or } 0.55}$ | M1 | Their $P(S)$ seen as denominator of fraction in the final answer, accept unsimplified |
| $\frac{177}{440}$ or $0.402$ | A1 | CAO |
| | **2** | |
---
## Question 7(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(0,1,2) = {^{10}C_0}(0.35)^0(0.65)^{10} + {^{10}C_1}(0.35)^1(0.65)^9 + {^{10}C_2}(0.35)^2(0.65)^8$ | M1 | One term: $^{10}C_x p^x(1-p)^{10-x}$ for $0 < x < 10$, any $0 < p < 1$ |
| $0.013463 + 0.072492 + 0.17565$ | A1 | Correct unsimplified expression, or better |
| $0.262$ | A1 | |
| | **3** | |
---
## Question 7(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 120 \times 0.35\ [= 42]$; Variance $= 120 \times 0.35 \times 0.65\ [= 27.3]$ | B1 | Correct mean and variance seen, allow unsimplified |
| $P(X > 32) = P\!\left(Z > \dfrac{32.5 - 42}{\sqrt{27.3}}\right) = P(Z > -1.818)$ | M1 | Substituting their mean and variance into $\pm$standardisation formula (any number), condone $\sigma^2$ or $\sqrt{\sigma}$ |
| | M1 | Using continuity correction $31.5$ or $32.5$ |
| $\Phi(1.818)$ | M1 | Appropriate area $\Phi$, from final process, must be probability |
| $0.966$ | A1 | $0.965 \leqslant p \leqslant 0.966$ |
| | **5** | |7 In the region of Arka, the total number of households in the three villages Reeta, Shan and Teber is 800 . Each of the households was asked about the quality of their broadband service. Their responses are summarised in the following table.
\begin{center}
\begin{tabular}{ | c | l | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{Quality of broadband service} \\
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & Excellent & Good & Poor \\
\hline
\multirow{3}{*}{Village} & Reeta & 75 & 118 & 32 \\
\cline { 2 - 5 }
& Shan & 223 & 177 & 40 \\
\cline { 2 - 5 }
& Teber & 12 & 60 & 63 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that a randomly chosen household is in Shan and has poor broadband service.
\item Find the probability that a randomly chosen household has good broadband service given that the household is in Shan.\\
In the whole of Arka there are a large number of households. A survey showed that $35 \%$ of households in Arka have no broadband service.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item 10 households in Arka are chosen at random.
Find the probability that fewer than 3 of these households have no broadband service.
\item 120 households in Arka are chosen at random.
Use an approximation to find the probability that more than 32 of these households have no broadband service.\\
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}\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q7 [11]}}