| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward application of the law of total probability and Bayes' theorem with clearly stated probabilities and no algebraic manipulation required. Part (a) is a direct tree diagram calculation, and part (b) is standard conditional probability using given values—both are textbook exercises requiring only careful arithmetic, making this easier than average. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2[\times1] + 0.45\times0.4 + 0.35\times0.3\) | M1 | \(0.2[\times1] + 0.45\times b + 0.35\times c\), \(b=0.4\), \(0.6\) \(c=0.3\), \(0.7\) |
| \(0.485\) or \(\frac{97}{200}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y\ | \bar{H}) = \frac{P(Y\cap\bar{H})}{P(\bar{H})} = \frac{0.35\times0.7}{1-their(\mathbf{a})} = \frac{0.245}{0.515}\) | B1 |
| M1 | \(0.515\) or \(1-their(\mathbf{a})\) or \([0.3\times0+]0.45\times d + 0.35\times e\), where \(d=their\ b'\), \(e=their\ c'\) seen as denominator of fraction | |
| \(0.476\) or \(\frac{49}{103}\) | A1 | \(0.4757 \leqslant p \leqslant 0.476\) |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2[\times1] + 0.45\times0.4 + 0.35\times0.3$ | M1 | $0.2[\times1] + 0.45\times b + 0.35\times c$, $b=0.4$, $0.6$ $c=0.3$, $0.7$ |
| $0.485$ or $\frac{97}{200}$ | A1 | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y\|\bar{H}) = \frac{P(Y\cap\bar{H})}{P(\bar{H})} = \frac{0.35\times0.7}{1-their(\mathbf{a})} = \frac{0.245}{0.515}$ | B1 | $0.35\times0.7$ or $0.245$ seen as numerator or denominator of fraction |
| | M1 | $0.515$ or $1-their(\mathbf{a})$ or $[0.3\times0+]0.45\times d + 0.35\times e$, where $d=their\ b'$, $e=their\ c'$ seen as denominator of fraction |
| $0.476$ or $\frac{49}{103}$ | A1 | $0.4757 \leqslant p \leqslant 0.476$ |2 Georgie has a red scarf, a blue scarf and a yellow scarf. Each day she wears exactly one of these scarves. The probabilities for the three colours are $0.2,0.45$ and 0.35 respectively. When she wears a red scarf, she always wears a hat. When she wears a blue scarf, she wears a hat with probability 0.4 . When she wears a yellow scarf, she wears a hat with probability 0.3 .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that on a randomly chosen day Georgie wears a hat.
\item Find the probability that on a randomly chosen day Georgie wears a yellow scarf given that she does not wear a hat.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q2 [5]}}