| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Person selected from combined populations |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using basic probability rules and Bayes' theorem. All required information is explicitly given, requiring only careful organization of data and application of standard formulas (P(A), P(no sugar) = weighted average, P(B|no sugar) using Bayes). No conceptual insight needed beyond recognizing the setup as a standard two-population problem. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(A) = 0.2\) | B1 1 | o.e. Must be single fraction or 20% |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{not } S) = 0.2 \times 0.7 + 0.8 \times 0.35\) | M1 | Summing two 2-factor probabilities or subtracting \(P(S)\) from 1 |
| \(= 0.42\) | A1 2 | o.e. Correct answer no decimals in fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(B \mid S') = \dfrac{0.8 \times 0.35}{0.42}\) | M1 | \(\dfrac{(1-\text{their}(i)) \times 0.35}{\text{their}(ii)}\) if marks lost in (i) or (ii) |
| \(= 0.667\) | A1 2 | Correct answer c.w.o |
## Question 2:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(A) = 0.2$ | B1 **1** | o.e. Must be single fraction or 20% |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{not } S) = 0.2 \times 0.7 + 0.8 \times 0.35$ | M1 | Summing two 2-factor probabilities or subtracting $P(S)$ from 1 |
| $= 0.42$ | A1 **2** | o.e. Correct answer no decimals in fractions |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(B \mid S') = \dfrac{0.8 \times 0.35}{0.42}$ | M1 | $\dfrac{(1-\text{their}(i)) \times 0.35}{\text{their}(ii)}$ if marks lost in (i) or (ii) |
| $= 0.667$ | A1 **2** | Correct answer c.w.o |
---2 In country $A 30 \%$ of people who drink tea have sugar in it. In country $B 65 \%$ of people who drink tea have sugar in it. There are 3 million people in country $A$ who drink tea and 12 million people in country $B$ who drink tea. A person is chosen at random from these 15 million people.\\
(i) Find the probability that the person chosen is from country $A$.\\
(ii) Find the probability that the person chosen does not have sugar in their tea.\\
(iii) Given that the person chosen does not have sugar in their tea, find the probability that the person is from country $B$.
\hfill \mbox{\textit{CAIE S1 2008 Q2 [5]}}