| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 3 |
| 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 question requiring basic probability calculations (counting and dividing) and checking independence using P(A∩B) = P(A)×P(B). The table is clearly laid out, arithmetic is simple, and the independence test is a standard procedure with no conceptual subtlety—easier than average A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Guitar | Piano | Flute | |
| Female students | 62 | 35 | 43 |
| Male students | 78 | 40 | 42 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{120}{300} = 0.4\) | B1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{male}) \times P(\text{not piano}) = \frac{160}{300} \times \frac{225}{300} \left(\frac{8}{15} \times \frac{3}{4}\right) = \frac{2}{5}\) | M1 | \(P(M) \times P(P')\) seen; can be unsimplified but events must be named in a product |
| As \(P(\text{male} \cap \text{not piano})\) also \(= \frac{120}{300} = \frac{2}{5}\); The events are Independent | A1 | Numerical comparison and correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{male} \cap \text{not piano}) = \frac{120}{300}\); \(P(\text{not piano}) = \frac{225}{300}\) | M1 | \(P(M |
| Answer | Marks | Guidance |
|---|---|---|
| Therefore the events are Independent | A1 | Numerical comparison with \(P(M)\) or \(P(P')\) and correct conclusion |
## Question 1:
### Part (i)
$\frac{120}{300} = 0.4$ | **B1** | OE
---
### Part (ii)
**Method 1:**
$P(\text{male}) \times P(\text{not piano}) = \frac{160}{300} \times \frac{225}{300} \left(\frac{8}{15} \times \frac{3}{4}\right) = \frac{2}{5}$ | **M1** | $P(M) \times P(P')$ seen; can be unsimplified but events must be named in a product
As $P(\text{male} \cap \text{not piano})$ also $= \frac{120}{300} = \frac{2}{5}$; The events are **Independent** | **A1** | Numerical comparison and correct conclusion
**Alternative Method:**
$P(\text{male} \cap \text{not piano}) = \frac{120}{300}$; $P(\text{not piano}) = \frac{225}{300}$ | **M1** | $P(M|P')$ or $P(P'|M)$ unsimplified seen with *their* probs with correctly named events
$P(M \mid \text{not piano}) = \frac{\frac{120}{300}}{\frac{225}{300}} = \frac{120}{225} = \frac{8}{15} = P(\text{male})$
**or**
$P(\text{not piano} \mid M) = \frac{\frac{120}{300}}{\frac{160}{300}} = \frac{120}{160} = \frac{3}{4} = P(\text{not piano})$
Therefore the events are **Independent** | **A1** | Numerical comparison with $P(M)$ or $P(P')$ and correct conclusion
---1 There are 300 students at a music college. All students play exactly one of the guitar, the piano or the flute. The numbers of male and female students that play each of the instruments are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Guitar & Piano & Flute \\
\hline
Female students & 62 & 35 & 43 \\
\hline
Male students & 78 & 40 & 42 \\
\hline
\end{tabular}
\end{center}
(i) Find the probability that a randomly chosen student at the college is a male who does not play the piano.\\
(ii) Determine whether the events 'a randomly chosen student is male' and 'a randomly chosen student does not play the piano' are independent, justifying your answer.\\
\hfill \mbox{\textit{CAIE S1 2019 Q1 [3]}}