| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Independence in contingency tables |
| Difficulty | Moderate -0.8 This is a straightforward contingency table completion followed by a routine independence check using P(X∩Y) = P(X)×P(Y). The arithmetic is simple (filling in 6, 19, 25 etc.) and the independence test is a standard procedure requiring only basic probability calculations with no conceptual challenges. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Wears spectacles | Does not wear spectacles | Total | |
| Right-handed | |||
| Not right-handed | |||
| Total | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with RH row: 6, 19, 25 and Not RH row: 2, 3, 5; Total row: 8, 22 | B1 | One correct row or column including total, other than the Total row/column |
| All values correct | B1 [2] | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X) = 25/30\), \(P(Y) = 8/30\) | M1 | \(P(X)\) or \(P(Y)\) from their table or correct from question (denom 30) |
| \(P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}\) | M1 | Comparing their \(P(X) \times P(Y)\) (values substituted) with their evaluated \(P(X \cap Y)\) – not \(P(X) \times P(Y)\) |
| \(P(X \cap Y) = 6/30 = 1/5 \neq P(X) \times P(Y)\) | ||
| Not independent | A1 [3] |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with RH row: 6, 19, 25 and Not RH row: 2, 3, 5; Total row: 8, 22 | **B1** | One correct row or column including total, other than the Total row/column |
| All values correct | **B1** [2] | All correct |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X) = 25/30$, $P(Y) = 8/30$ | **M1** | $P(X)$ or $P(Y)$ from their table or correct from question (denom 30) |
| $P(X) \times P(Y) = \frac{25}{30} \times \frac{8}{30} = \frac{200}{900} = \frac{2}{9}$ | **M1** | Comparing their $P(X) \times P(Y)$ (values substituted) with their evaluated $P(X \cap Y)$ – not $P(X) \times P(Y)$ |
| $P(X \cap Y) = 6/30 = 1/5 \neq P(X) \times P(Y)$ | | |
| Not independent | **A1** [3] | |
---1 In a group of 30 adults, 25 are right-handed and 8 wear spectacles. The number who are right-handed and do not wear spectacles is 19 .\\
(i) Copy and complete the following table to show the number of adults in each category.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Wears spectacles & Does not wear spectacles & Total \\
\hline
Right-handed & & & \\
\hline
Not right-handed & & & \\
\hline
Total & & & 30 \\
\hline
\end{tabular}
\end{center}
An adult is chosen at random from the group. Event $X$ is 'the adult chosen is right-handed'; event $Y$ is 'the adult chosen wears spectacles'.\\
(ii) Determine whether $X$ and $Y$ are independent events, justifying your answer.
\hfill \mbox{\textit{CAIE S1 2016 Q1 [5]}}