| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Independence in contingency tables |
| Difficulty | Moderate -0.3 This is a straightforward contingency table question testing basic probability and independence concepts. Part (a) requires simple probability calculation and checking independence using P(A∩B)=P(A)P(B), which is standard S1 material. Parts (b)(i) and (b)(ii) are routine binomial and normal approximation applications with no conceptual challenges—slightly easier than average A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Swimming | Cycling | Running | |
| Female | 104 | 50 | 66 |
| Male | 31 | 57 | 92 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\frac{104+31}{400}\right] = \frac{135}{400}, \frac{27}{80}, 0.3375\) | B1 | Evaluated, exact value |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(M) = \frac{180}{400}\), \(P(S) = \frac{135}{400}\), \(P(M \cap S) = \frac{31}{400}\); \(\frac{180}{400} \times \frac{135}{400} = \frac{243}{1600}\), \(0.151875 \neq \frac{31}{400}\) so NOT independent | M1 | *Their* \(P(M) \times\) *their* \(P(S)\) seen, accept unsimplified |
| A1 | \(P(M)\), \(P(S)\) and \(P(M \cap S)\) notation seen, numerical comparison and correct conclusion, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(M \cap S) = \frac{31}{400}\), \(P(S) = \frac{135}{400}\), \(P(M) = \frac{180}{400}\); \(P(M | S) = \frac{\frac{31}{400}}{\frac{135}{400}} = \frac{31}{135}\), \(0.2296\ldots \neq \frac{180}{400}\) so NOT independent | M1 |
| A1 | \(P(M)\), \(P(S)\) and \(P(M \cap S)\) notation seen, numerical comparison and correct conclusion, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1 - ({}^{10}C_0 \, 0.3^0 \, 0.7^{10} + {}^{10}C_1 \, 0.3^1 \, 0.7^9 + {}^{10}C_2 \, 0.3^2 \, 0.7^8)\) | M1 | \({}^{10}C_x \, p^x (1-p)^{10-x}\) for \(0 < x < 10\), \(0 < p < 1\), any \(p\) |
| \(= 1 - (0.028248 + 0.121061 + 0.233474)\) | A1 | Correct expression, accept unsimplified, condone omission of final bracket, condone recovery from poor notation |
| \(= 0.617\) | A1 | Accept \(0.61715 \leq p \leq 0.61722\), WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \({}^{10}C_3 \, 0.3^3 \, 0.7^7 + {}^{10}C_4 \, 0.3^4 \, 0.7^6 + {}^{10}C_5 \, 0.3^5 \, 0.7^5 + {}^{10}C_6 \, 0.3^6 \, 0.7^4 + {}^{10}C_7 \, 0.3^7 \, 0.7^3 + {}^{10}C_8 \, 0.3^8 \, 0.7^2 + {}^{10}C_9 \, 0.3^9 \, 0.7^1 + {}^{10}C_{10} \, 0.3^{10} \, 0.7^0\) | M1 | \({}^{10}C_x \, p^x (1-p)^{10-x}\) for \(0 < x < 10\), \(0 < p < 1\), any \(p\) |
| A1 | Correct unsimplified expression | |
| \(= 0.617\) | A1 | Accept \(0.61715 \leq p \leq 0.61722\), WWW |
| Answer | Marks | Guidance |
|---|---|---|
| \([p = 0.3]\) Mean \(= 0.3 \times 90 = 27\); variance \(= 0.3 \times 90 \times 0.7 = 18.9\) | B1 | Correct mean and variance, allow unsimplified. Condone \(\sigma = 4.347\) evaluated |
| \(P(X < 32) = P\!\left(z < \frac{31.5 - 27}{\sqrt{18.9}}\right)\) | M1 | Substituting *their* \(\mu\) and \(\sigma\) (not \(\sigma^2\), \(\sqrt{\sigma}\)) into the \(\pm\)standardising formula with a numerical value for \(31.5\) |
| M1 | Using either \(31.5\) or \(32.5\) within a \(\pm\)standardising formula with numerical values for *their* \(\mu\) and \(\sigma\) (condone \(\sigma^2\), \(\sqrt{\sigma}\)) | |
| \(= \Phi(1.035)\) | M1 | Appropriate area \(\Phi\), from standardisation formula \(P(z<\ldots)\) in final solution, must be probability |
| \(= 0.850\) | A1 | Allow \(0.8495 < p \leq 0.85(0)\), final answer WWW |
## Question 7(a)(i):
| $\left[\frac{104+31}{400}\right] = \frac{135}{400}, \frac{27}{80}, 0.3375$ | **B1** | Evaluated, exact value |
---
## Question 7(a)(ii):
**Method 1:**
| $P(M) = \frac{180}{400}$, $P(S) = \frac{135}{400}$, $P(M \cap S) = \frac{31}{400}$; $\frac{180}{400} \times \frac{135}{400} = \frac{243}{1600}$, $0.151875 \neq \frac{31}{400}$ so NOT independent | **M1** | *Their* $P(M) \times$ *their* $P(S)$ seen, accept unsimplified |
|---|---|---|
| | **A1** | $P(M)$, $P(S)$ and $P(M \cap S)$ notation seen, numerical comparison and correct conclusion, WWW |
**Method 2:**
| $P(M \cap S) = \frac{31}{400}$, $P(S) = \frac{135}{400}$, $P(M) = \frac{180}{400}$; $P(M|S) = \frac{\frac{31}{400}}{\frac{135}{400}} = \frac{31}{135}$, $0.2296\ldots \neq \frac{180}{400}$ so NOT independent | **M1** | $[P(M|S) =] \frac{\textit{their } P(M \cap S)}{\textit{their } P(S)}$ (oe) seen, accept unsimplified |
|---|---|---|
| | **A1** | $P(M)$, $P(S)$ and $P(M \cap S)$ notation seen, numerical comparison and correct conclusion, WWW |
---
## Question 7(b)(i):
**Method 1** $[1 - P(0,1,2)]$:
| $= 1 - ({}^{10}C_0 \, 0.3^0 \, 0.7^{10} + {}^{10}C_1 \, 0.3^1 \, 0.7^9 + {}^{10}C_2 \, 0.3^2 \, 0.7^8)$ | **M1** | ${}^{10}C_x \, p^x (1-p)^{10-x}$ for $0 < x < 10$, $0 < p < 1$, any $p$ |
|---|---|---|
| $= 1 - (0.028248 + 0.121061 + 0.233474)$ | **A1** | Correct expression, accept unsimplified, condone omission of final bracket, condone recovery from poor notation |
| $= 0.617$ | **A1** | Accept $0.61715 \leq p \leq 0.61722$, WWW |
**Method 2** $[P(3,4,5,6,7,8,9,10) =]$:
| ${}^{10}C_3 \, 0.3^3 \, 0.7^7 + {}^{10}C_4 \, 0.3^4 \, 0.7^6 + {}^{10}C_5 \, 0.3^5 \, 0.7^5 + {}^{10}C_6 \, 0.3^6 \, 0.7^4 + {}^{10}C_7 \, 0.3^7 \, 0.7^3 + {}^{10}C_8 \, 0.3^8 \, 0.7^2 + {}^{10}C_9 \, 0.3^9 \, 0.7^1 + {}^{10}C_{10} \, 0.3^{10} \, 0.7^0$ | **M1** | ${}^{10}C_x \, p^x (1-p)^{10-x}$ for $0 < x < 10$, $0 < p < 1$, any $p$ |
|---|---|---|
| | **A1** | Correct unsimplified expression |
| $= 0.617$ | **A1** | Accept $0.61715 \leq p \leq 0.61722$, WWW |
---
## Question 7(b)(ii):
| $[p = 0.3]$ Mean $= 0.3 \times 90 = 27$; variance $= 0.3 \times 90 \times 0.7 = 18.9$ | **B1** | Correct mean and variance, allow unsimplified. Condone $\sigma = 4.347$ evaluated |
|---|---|---|
| $P(X < 32) = P\!\left(z < \frac{31.5 - 27}{\sqrt{18.9}}\right)$ | **M1** | Substituting *their* $\mu$ and $\sigma$ (not $\sigma^2$, $\sqrt{\sigma}$) into the $\pm$standardising formula with a numerical value for $31.5$ |
| | **M1** | Using either $31.5$ or $32.5$ within a $\pm$standardising formula with numerical values for *their* $\mu$ and $\sigma$ (condone $\sigma^2$, $\sqrt{\sigma}$) |
| $= \Phi(1.035)$ | **M1** | Appropriate area $\Phi$, from standardisation formula $P(z<\ldots)$ in final solution, must be probability |
| $= 0.850$ | **A1** | Allow $0.8495 < p \leq 0.85(0)$, final answer WWW |7 There are 400 students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the results are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Swimming & Cycling & Running \\
\hline
Female & 104 & 50 & 66 \\
\hline
Male & 31 & 57 & 92 \\
\hline
\end{tabular}
\end{center}
A student is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the probability that the student prefers swimming.
\item Determine whether the events 'the student is male' and 'the student prefers swimming' are independent, justifying your answer.\\
On average at all the schools in this country $30 \%$ of the students do not like any sports.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item 10 of the students from this country are chosen at random.
Find the probability that at least 3 of these students do not like any sports.
\item 90 students from this country are now chosen at random.
Use an approximation to find the probability that fewer than 32 of them do not like any sports.\\
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]}}