| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| 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 (a) requires setting up one equation with one unknown using P(not late) = 0.48, which is algebraically simple. Part (b) is a direct Bayes' theorem calculation. Both parts are standard textbook exercises requiring no novel insight, making this slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{not late}) = 0.4 \times 0.45 + 0.35 \times 0.3 + 0.25 \times (1-x)\) or \(P(\text{late}) = 0.4 \times 0.55 + 0.35 \times 0.7 + 0.25x\) | M1 | \(0.4 \times p + 0.35 \times q + 0.25 \times r\), \(p = 0.45, 0.55\), \(q = 0.3, 0.7\) and \(r = (1-x), x\) |
| \(0.18 + 0.105 + 0.25(1-x) = 0.48\) or \(0.22 + 0.245 + 0.25x = 0.52\) | A1 | Linear equation formed using sum of 3 probabilities and 0.48 or 0.52 as appropriate. Accept unsimplified. |
| \(x = 0.22\) | A1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[P(\text{train} \mid \text{late}) = \frac{P(\text{train} \cap \text{late})}{P(\text{late})}\right]\) | B1 | \(0.35 \times 0.7\) or \(0.245\) seen as numerator of fraction |
| \(= \frac{0.35 \times 0.7}{1 - 0.48}\) or \(\frac{0.35 \times 0.7}{0.4 \times 0.55 + 0.35 \times 0.7 + 0.25 \times \textit{their } 0.22}\) | M1 | \(P(\text{late})\) seen as denominator with *their* probability as numerator. Accept \(\frac{\textit{their } p}{0.52}\) or \(\frac{\textit{their } p}{0.22 + 0.245 + 0.25 \times \textit{their } 0.22}\) |
| \(= 0.471\) or \(\frac{49}{104}\) | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{not late}) = 0.4 \times 0.45 + 0.35 \times 0.3 + 0.25 \times (1-x)$ or $P(\text{late}) = 0.4 \times 0.55 + 0.35 \times 0.7 + 0.25x$ | M1 | $0.4 \times p + 0.35 \times q + 0.25 \times r$, $p = 0.45, 0.55$, $q = 0.3, 0.7$ and $r = (1-x), x$ |
| $0.18 + 0.105 + 0.25(1-x) = 0.48$ or $0.22 + 0.245 + 0.25x = 0.52$ | A1 | Linear equation formed using sum of 3 probabilities and 0.48 or 0.52 as appropriate. Accept unsimplified. |
| $x = 0.22$ | A1 | Final answer |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[P(\text{train} \mid \text{late}) = \frac{P(\text{train} \cap \text{late})}{P(\text{late})}\right]$ | B1 | $0.35 \times 0.7$ or $0.245$ seen as numerator of fraction |
| $= \frac{0.35 \times 0.7}{1 - 0.48}$ or $\frac{0.35 \times 0.7}{0.4 \times 0.55 + 0.35 \times 0.7 + 0.25 \times \textit{their } 0.22}$ | M1 | $P(\text{late})$ seen as denominator with *their* probability as numerator. Accept $\frac{\textit{their } p}{0.52}$ or $\frac{\textit{their } p}{0.22 + 0.245 + 0.25 \times \textit{their } 0.22}$ |
| $= 0.471$ or $\frac{49}{104}$ | A1 | |
---3 On each day that Alexa goes to work, the probabilities that she travels by bus, by train or by car are $0.4,0.35$ and 0.25 respectively. When she travels by bus, the probability that she arrives late is 0.55 . When she travels by train, the probability that she arrives late is 0.7 . When she travels by car, the probability that she arrives late is $x$.
On a randomly chosen day when Alexa goes to work, the probability that she does not arrive late is 0.48 .
\begin{enumerate}[label=(\alph*)]
\item Find the value of $x$.
\item Find the probability that Alexa travels to work by train given that she arrives late.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q3 [6]}}