| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Two-way table probabilities |
| Difficulty | Easy -1.8 This is a straightforward two-way table question requiring only basic probability calculations: (a) simple probability from totals, (b) direct conditional probability, and (c) checking independence using P(A∩B)=P(A)P(B). All values are given in the table with no problem-solving or insight required—purely mechanical arithmetic. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Piano | Guitar | Drums | |
| Male | 25 | 44 | 11 |
| Female | 42 | 38 | 20 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{82}{180}, \frac{41}{90}\), 0.456 | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[P(M\ | D) = \frac{P(M \cap D)}{P(D)}\right] = \frac{\frac{11}{180}}{\frac{20}{180}+\frac{11}{180}}\) or \(\frac{0.6011}{0.1722}\) | M1 |
| \(\frac{11}{31}\), 0.355 | A1 | Final answer |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F) = \frac{100}{180}, \frac{5}{9}, 0.5556\) OE; \(P(G) = \frac{82}{180}, \frac{41}{90}, 0.4556\) OE | M1 | *Their* identified \(P(F) \times\) *their* identified \(P(G)\) or correct seen, can be unsimplified |
| \(P(F \cap G) = \frac{38}{180}, \frac{19}{90}, 0.2111\) OE; \(P(F) \times P(G) = \frac{100}{180} \times \frac{82}{180} = \frac{41}{162}, 0.2531\) OE \(\left[\neq \frac{38}{180}\right]\); Not independent | A1 | \(\frac{41}{162}, \frac{38}{180}\), \(P(F \cap G)\) and \(P(F) \times P(G)\) seen with correct conclusion, WWW. Values and labels must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F \cap G) = \frac{38}{180}, \frac{19}{90}, 0.2111\) OE; \(P(G) = \frac{82}{180}, \frac{41}{90}, 0.4556\) OE | M1 | \(P(F\ |
| \(P(F\ | G) = \dfrac{\frac{38}{180}}{\frac{82}{180}} = \frac{19}{41}, 0.4634\) OE; \(\neq P(F) = \frac{100}{180}, \frac{5}{9}, 0.5556\) OE; Not independent | A1 |
## Question 1:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{82}{180}, \frac{41}{90}$, 0.456 | B1 | |
| | **Total: 1** | |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[P(M\|D) = \frac{P(M \cap D)}{P(D)}\right] = \frac{\frac{11}{180}}{\frac{20}{180}+\frac{11}{180}}$ or $\frac{0.6011}{0.1722}$ | M1 | Their identified $\frac{P(M \cap D)}{P(D)}$, or from data table $\frac{11}{20+11}$, accept unsimplified, condone $\times 180$ |
| $\frac{11}{31}$, 0.355 | A1 | Final answer |
| | **Total: 2** | |
## Question 1(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F) = \frac{100}{180}, \frac{5}{9}, 0.5556$ OE; $P(G) = \frac{82}{180}, \frac{41}{90}, 0.4556$ OE | M1 | *Their* identified $P(F) \times$ *their* identified $P(G)$ or correct seen, can be unsimplified |
| $P(F \cap G) = \frac{38}{180}, \frac{19}{90}, 0.2111$ OE; $P(F) \times P(G) = \frac{100}{180} \times \frac{82}{180} = \frac{41}{162}, 0.2531$ OE $\left[\neq \frac{38}{180}\right]$; Not independent | A1 | $\frac{41}{162}, \frac{38}{180}$, $P(F \cap G)$ and $P(F) \times P(G)$ seen with correct conclusion, WWW. Values and labels must be seen |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F \cap G) = \frac{38}{180}, \frac{19}{90}, 0.2111$ OE; $P(G) = \frac{82}{180}, \frac{41}{90}, 0.4556$ OE | M1 | $P(F\|G)$ (OE) unsimplified with *their* identified probs or correct |
| $P(F\|G) = \dfrac{\frac{38}{180}}{\frac{82}{180}} = \frac{19}{41}, 0.4634$ OE; $\neq P(F) = \frac{100}{180}, \frac{5}{9}, 0.5556$ OE; Not independent | A1 | $\frac{19}{41}, \frac{100}{180}$, $P(F \cap G)$ and $P(F\|G)$ seen with correct conclusion WWW. Values and labels must be seen |
**Total: 2 marks**
---1 Each of the 180 students at a college plays exactly one of the piano, the guitar and the drums. The numbers of male and female students who play the piano, the guitar and the drums are given in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Piano & Guitar & Drums \\
\hline
Male & 25 & 44 & 11 \\
\hline
Female & 42 & 38 & 20 \\
\hline
\end{tabular}
\end{center}
A student at the college is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the student plays the guitar.
\item Find the probability that the student is male given that the student plays the drums.
\item Determine whether the events 'the student plays the guitar' and 'the student is female' are independent, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q1 [5]}}