| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Moderate -0.3 This is a straightforward application of Bayes' theorem with clearly presented data in a table. Students must calculate P(Mumbok|18-60) using conditional probability, requiring organization of given information and basic arithmetic with probabilities. The structure is standard for S1 conditional probability questions, slightly easier than average A-level due to clear setup and routine method, though the multi-step calculation prevents it from being trivial. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Mumbok | Bagville | |
| Under 18 years | 15 | 35 |
| 18 to 60 years | 55 | 95 |
| Over 60 years | 20 | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| either \(\frac{55}{90}\) \((\frac{11}{18})\) or \(\frac{95}{160}\) \((\frac{19}{32})\) seen | B1 | oe |
| \(P(M \text{ and } 18-60) = 0.6 \times \frac{55}{90} = 0.367\) \((\frac{11}{30})\) | M1 | 0.6 mult by \(\frac{55}{90}\) seen as num/denom of a fraction |
| \(P(18-60) = 0.6 \times \frac{55}{90} + 0.4 \times \frac{95}{160}\) \((= \frac{29}{48}\) or \(0.604)\) | M1 | Summing 2 two-factor products seen anywhere |
| \(P(M \mid 18-60) = \dfrac{P(M \cap 18-60)}{P(18-60)}\) | A1 | Correct unsimplified answer seen as num/denom of a fraction |
| \(= \frac{88}{145}\) \((0.607)\) | A1 [5] | Correct answer |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| either $\frac{55}{90}$ $(\frac{11}{18})$ or $\frac{95}{160}$ $(\frac{19}{32})$ seen | B1 | oe |
| $P(M \text{ and } 18-60) = 0.6 \times \frac{55}{90} = 0.367$ $(\frac{11}{30})$ | M1 | 0.6 mult by $\frac{55}{90}$ seen as num/denom of a fraction |
| $P(18-60) = 0.6 \times \frac{55}{90} + 0.4 \times \frac{95}{160}$ $(= \frac{29}{48}$ or $0.604)$ | M1 | Summing 2 two-factor products seen anywhere |
| $P(M \mid 18-60) = \dfrac{P(M \cap 18-60)}{P(18-60)}$ | A1 | Correct unsimplified answer seen as num/denom of a fraction |
| $= \frac{88}{145}$ $(0.607)$ | A1 **[5]** | Correct answer |
---2 The people living in two towns, Mumbok and Bagville, are classified by age. The numbers in thousands living in each town are shown in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Mumbok & Bagville \\
\hline
Under 18 years & 15 & 35 \\
\hline
18 to 60 years & 55 & 95 \\
\hline
Over 60 years & 20 & 30 \\
\hline
\end{tabular}
\end{center}
One of the towns is chosen. The probability of choosing Mumbok is 0.6 and the probability of choosing Bagville is 0.4 . Then a person is chosen at random from that town. Given that the person chosen is between 18 and 60 years old, find the probability that the town chosen was Mumbok. [5]
\hfill \mbox{\textit{CAIE S1 2013 Q2 [5]}}