| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.3 This is a straightforward two-way table question requiring only basic probability calculations (reading from table, dividing by total) and a simple independence check using the definition P(M∩E) = P(M)×P(E). All values are given directly in the table with no problem-solving or conceptual insight needed—purely mechanical application of formulas taught in early statistics. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | Unemployed | Employed |
| Male | 206 | 412 |
| Female | 358 | 305 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For correct numerator | B1 | |
| For correct denominator | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Follow through on their denominator if \(p < 1\) or \(\frac{2}{3} \times\) their (i) | B1ft [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For attempting to find \(P(E)\) | M1 | |
| For showing they know what independence means, mathematically | M1dep | |
| ft on their (i) \(\times\) their \(P(E) \neq\) their (ii) | A1ft [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For correct numerator, 0.28 gets B0 with PA | B1 | |
| For correct denominator | B1 [2] |
(i) $\frac{618}{1281}$ (0.482)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For correct numerator | B1 | |
| For correct denominator | B1 [2] | |
(ii) $\frac{412}{1281}$ (0.322) or tree diagram options
| Answer | Marks | Guidance |
|--------|-------|----------|
| Follow through on their denominator if $p < 1$ or $\frac{2}{3} \times$ their (i) | B1ft [1] | |
(iii) $P(E) = \frac{717}{1281}$
Their (i) $\times$ their $P(E) \neq$ their (ii)
Not independent
| Answer | Marks | Guidance |
|--------|-------|----------|
| For attempting to find $P(E)$ | M1 | |
| For showing they know what independence means, mathematically | M1dep | |
| ft on their (i) $\times$ their $P(E) \neq$ their (ii) | A1ft [3] | |
(iv) $\frac{358}{564}$ (0.635) or (0.279/0.440)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For correct numerator, 0.28 gets B0 with PA | B1 | |
| For correct denominator | B1 [2] | |
---5 Data about employment for males and females in a small rural area are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & Unemployed & Employed \\
\hline
Male & 206 & 412 \\
\hline
Female & 358 & 305 \\
\hline
\end{tabular}
\end{center}
A person from this area is chosen at random. Let $M$ be the event that the person is male and let $E$ be the event that the person is employed.\\
(i) Find $\mathrm { P } ( M )$.\\
(ii) Find $\mathrm { P } ( M$ and $E )$.\\
(iii) Are $M$ and $E$ independent events? Justify your answer.\\
(iv) Given that the person chosen is unemployed, find the probability that the person is female.
\hfill \mbox{\textit{CAIE S1 2005 Q5 [8]}}