| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Finding unknown probability from total probability |
| Difficulty | Moderate -0.3 This is a straightforward application of the law of total probability and Bayes' theorem. Part (i) requires setting up and solving a simple linear equation (0.4x + 0.6(2x) = 0.36), while part (ii) is a standard conditional probability calculation using Bayes' theorem. Both parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4x + 0.6 \times 2x = 0.36\) or \(0.4(1-x) + 0.6(1-2x) = 0.64\) | M1 | \(0.4a + (1-0.4)b = 0.36\) or \(0.64\), \(a,b\) terms involving \(x\) |
| \(1.6x = 0.36\), \(x = 0.225\) | A1 | Fully justified by algebra; AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(H\ | L') = \dfrac{0.4(1-x)}{1-0.36} = \dfrac{0.4\times(1-0.225)}{0.64} = \dfrac{0.4\times 0.775}{0.4\times 0.775 + 0.6\times 0.55}\) | M1 |
| M1 | Denominator \(0.36\) or \(0.64\). Allow unsimplified. | |
| \(\dfrac{31}{64}\) or \(0.484\) | A1 | |
| Total: 3 |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4x + 0.6 \times 2x = 0.36$ or $0.4(1-x) + 0.6(1-2x) = 0.64$ | **M1** | $0.4a + (1-0.4)b = 0.36$ or $0.64$, $a,b$ terms involving $x$ |
| $1.6x = 0.36$, $x = 0.225$ | **A1** | Fully justified by algebra; AG |
| **Total: 2** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(H\|L') = \dfrac{0.4(1-x)}{1-0.36} = \dfrac{0.4\times(1-0.225)}{0.64} = \dfrac{0.4\times 0.775}{0.4\times 0.775 + 0.6\times 0.55}$ | **M1** | Correct numerical numerator of a fraction. Allow unsimplified. |
| | **M1** | Denominator $0.36$ or $0.64$. Allow unsimplified. |
| $\dfrac{31}{64}$ or $0.484$ | **A1** | |
| **Total: 3** | | |
---2 Benju cycles to work each morning and he has two possible routes. He chooses the hilly route with probability 0.4 and the busy route with probability 0.6 . If he chooses the hilly route, the probability that he will be late for work is $x$ and if he chooses the busy route the probability that he will be late for work is $2 x$. The probability that Benju is late for work on any day is 0.36 .\\
(i) Show that $x = 0.225$.\\
(ii) Given that Benju is not late for work, find the probability that he chooses the hilly route.\\
\hfill \mbox{\textit{CAIE S1 2019 Q2 [5]}}