| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Markov chain weather transitions |
| Difficulty | Moderate -0.3 This is a straightforward Markov chain problem requiring tree diagram completion, basic probability calculations using the law of total probability, and conditional probability via Bayes' theorem. While it involves multiple parts and extends to three days, each step follows standard AS-level procedures without requiring novel insight or complex reasoning. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram with probabilities: 1 April Fine \(= 0.8\), Rainy \(= 0.2\); 2 April given Fine: Fine \(= 0.75\), Rainy \(= 0.25\); 2 April given Rainy: Fine \(= 0.4\), Rainy \(= 0.6\) | B1 | All probabilities correct, may be on branch or next to 'Fine/Rainy'. Ignore additional branches. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.8 \times 0.75 + 0.2 \times 0.4\ (= 0.6 + 0.08)\) | M1 | Correct or FT from their diagram unsimplified, all probabilities \(0 < p < 1\). Partial evaluation only sufficient when correct. Accept working in 4(b) or by the tree diagram. |
| \(0.68,\ \frac{17}{25}\) | A1 | From supporting working |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.8 \times 0.75 \times 0.25 + 0.8 \times 0.25 \times 0.6\) | M1 | \(a \times b \times c + a \times 1{-}b \times d\), \(0 < c, d \leq 1\), \(a,b\) consistent with their tree diagram or correct, no additional terms |
| \(0.15 + 0.12\) | A1 | At least one term correct, accept unsimplified |
| \(0.27\) | A1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y) = \text{their (c)} + 0.2 \times 0.4 \times 0.25 + 0.2 \times 0.6 \times 0.6\ (= 0.362)\) | B1 FT | their \(\mathbf{(c)} + e \times f \times g + e \times (1{-}f) \times h\), \(0 < g, h \leq 1\), \(e,f\) consistent with their tree diagram or correct |
| \(P(X | Y) = \frac{\text{their (c)}}{\text{their } P(Y)} = \frac{0.27}{0.362}\) | M1 |
| \(0.746,\ \frac{373}{500}\) or \(\frac{135}{181}\) | A1 | \((0.7458\ldots)\) |
## Question 4(a):
Tree diagram with probabilities: 1 April Fine $= 0.8$, Rainy $= 0.2$; 2 April given Fine: Fine $= 0.75$, Rainy $= 0.25$; 2 April given Rainy: Fine $= 0.4$, Rainy $= 0.6$ | B1 | All probabilities correct, may be on branch or next to 'Fine/Rainy'. Ignore additional branches.
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## Question 4(b):
$0.8 \times 0.75 + 0.2 \times 0.4\ (= 0.6 + 0.08)$ | M1 | Correct or FT from their diagram unsimplified, all probabilities $0 < p < 1$. Partial evaluation only sufficient when correct. Accept working in **4(b)** or by the tree diagram.
$0.68,\ \frac{17}{25}$ | A1 | From supporting working
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## Question 4(c):
$0.8 \times 0.75 \times 0.25 + 0.8 \times 0.25 \times 0.6$ | M1 | $a \times b \times c + a \times 1{-}b \times d$, $0 < c, d \leq 1$, $a,b$ consistent with their tree diagram or correct, no additional terms
$0.15 + 0.12$ | A1 | At least one term correct, accept unsimplified
$0.27$ | A1 | Final answer
---
## Question 4(d):
$P(Y) = \text{their (c)} + 0.2 \times 0.4 \times 0.25 + 0.2 \times 0.6 \times 0.6\ (= 0.362)$ | B1 FT | their $\mathbf{(c)} + e \times f \times g + e \times (1{-}f) \times h$, $0 < g, h \leq 1$, $e,f$ consistent with their tree diagram or correct
$P(X|Y) = \frac{\text{their (c)}}{\text{their } P(Y)} = \frac{0.27}{0.362}$ | M1 | their **4(c)** (or correct)/their previously calculated and identified $P(Y)$ or a denominator involving 3 or 4 three-factor probability terms consistent with their tree diagram, third factor $0 < p < 1$
$0.746,\ \frac{373}{500}$ or $\frac{135}{181}$ | A1 | $(0.7458\ldots)$4 In a certain country, the weather each day is classified as fine or rainy. The probability that a fine day is followed by a fine day is 0.75 and the probability that a rainy day is followed by a fine day is 0.4 . The probability that it is fine on 1 April is 0.8 . The tree diagram below shows the possibilities for the weather on 1 April and 2 April.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram to show the probabilities.
1 April\\
\includegraphics[max width=\textwidth, alt={}, center]{33c0bd01-f27b-424c-a78a-6f36178bc08c-08_601_405_706_408}
2 April
Fine
Rainy
Fine
Rainy
\item Find the probability that 2 April is fine.\\
Let $X$ be the event that 1 April is fine and $Y$ be the event that 3 April is rainy.
\item Find the value of $\mathrm { P } ( X \cap Y )$.
\item Find the probability that 1 April is fine given that 3 April is rainy.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q4 [9]}}