| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Finding unknown probability from total probability |
| Difficulty | Standard +0.3 This is a straightforward application of the law of total probability to find x, followed by a standard conditional probability calculation using Bayes' theorem. The setup is clear, requires only algebraic manipulation of given probabilities, and involves no conceptual subtlety—slightly easier than average for A-level. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2 \times x + 0.1 \times 2x + 0.7 \times 0.25 = 0.235\) | M1 | \(0.2 \times x + 0.1 \times 2x + 0.7 \times 0.25\) or \(0.2x + 0.2x + 0.175\) seen |
| (equating their 3 term expression involving \(x\) to 0.235) | M1 | Equating *their* 3 term expression (2 terms involving \(x\)) to 0.235 |
| \(x = 0.15\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{car} \mid \text{not late}) = \dfrac{P(\text{car and not late})}{P(\text{not late})}\), \(\dfrac{0.1 \times (1 - 0.3)}{1 - 0.235}\) | M1 | \(0.1 \times (1 - 2 \times \text{their } x)\) or \(0.1 \times 0.7\) as numerator and \(0.2 \times (1 - \text{their } x) + 0.1 \times (1 - 2 \times \text{their } x) + 0.7 \times 0.75\) with values substituted or \(1 - 0.235\) or \(0.765\) as denominator. Condone \(0.2 \times (1 - \text{their } x) + 0.1 \times (1 - \times \text{their } x) + 0.7 \times 0.75\) as denominator consistent with 1(a) |
| \(\dfrac{0.07}{0.765} = 0.0915, \dfrac{70}{765}, \dfrac{14}{153}\) | A1 | 0.091503267 to at least 3SF. If M0 scored SC B1 for 0.091503267 to at least 3SF |
| Total: 2 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2 \times x + 0.1 \times 2x + 0.7 \times 0.25 = 0.235$ | M1 | $0.2 \times x + 0.1 \times 2x + 0.7 \times 0.25$ or $0.2x + 0.2x + 0.175$ seen |
| (equating their 3 term expression involving $x$ to 0.235) | M1 | Equating *their* 3 term expression (2 terms involving $x$) to 0.235 |
| $x = 0.15$ | A1 | |
| **Total: 3** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{car} \mid \text{not late}) = \dfrac{P(\text{car and not late})}{P(\text{not late})}$, $\dfrac{0.1 \times (1 - 0.3)}{1 - 0.235}$ | M1 | $0.1 \times (1 - 2 \times \text{their } x)$ or $0.1 \times 0.7$ as numerator **and** $0.2 \times (1 - \text{their } x) + 0.1 \times (1 - 2 \times \text{their } x) + 0.7 \times 0.75$ with values substituted or $1 - 0.235$ or $0.765$ as denominator. Condone $0.2 \times (1 - \text{their } x) + 0.1 \times (1 - \times \text{their } x) + 0.7 \times 0.75$ as denominator consistent with **1(a)** |
| $\dfrac{0.07}{0.765} = 0.0915, \dfrac{70}{765}, \dfrac{14}{153}$ | A1 | 0.091503267 to at least 3SF. If M0 scored **SC B1** for 0.091503267 to at least 3SF |
| **Total: 2** | | |1 On any day, Kino travels to school by bus, by car or on foot with probabilities 0.2, 0.1 and 0.7 respectively. The probability that he is late when he travels by bus is $x$. The probability that he is late when he travels by car is $2 x$ and the probability that he is late when he travels on foot is 0.25 .
The probability that, on a randomly chosen day, Kino is late is 0.235 .
\begin{enumerate}[label=(\alph*)]
\item Find the value of $x$.
\item Find the probability that, on a randomly chosen day, Kino travels to school by car given that he is not late.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q1 [5]}}