| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Both independence and mutual exclusivity |
| Difficulty | Standard +0.3 This is a straightforward probability question requiring students to list outcomes for two dice, calculate P(A), P(B), and P(A∩B), then check if P(A∩B) = P(A)×P(B). Part (b) uses basic conditional probability formula. While it requires careful enumeration and organization, it involves only standard S1 techniques with no novel insight, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(A) = \frac{10}{36}\), \(P(B) = \frac{24}{36}\) | B1 | Accept \(P(A) = \frac{10}{36}, \frac{5}{18}, 0.278\) and \(P(B) = \frac{24}{36}, \frac{2}{3}, 0.667\) |
| \(P(A \cap B) = \frac{8}{36}\) | B1 | |
| \(\frac{10}{36} \times \frac{24}{36}\) | M1 | Their \(P(A) \times\) their \(P(B)\) seen numerically, \(0 \leq\) their \(P(A), P(B) \leq 1\) |
| \(= \frac{5}{27}, 0.185 \left[\neq \frac{8}{36}\right]\); Events are not independent | A1 FT | Multiplication evaluated correctly and compared with intersection that is not a product of multiplication, conclusion stated, notation \(P(A)\), \(P(B)\) and \(P(A\cap B)\) used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[P(B\ | A') = \frac{P(B\cap A')}{P(A')} = \right] \frac{\frac{16}{36}}{\left(1-\frac{10}{36}\right)}\) | M1 |
| M1 | \([P(A') =]\left(1-\frac{10}{36}\right), \frac{26}{36}, 0.7222\) or \(1-\) their \(P(A)\) seen as denominator of conditional probability fraction | |
| \(= \frac{8}{13}\) | A1 | Final answer \(\frac{16}{26}, \frac{8}{13}, 0.6153846\) to at least 3SF |
| Alternative (direct from outcome tables): \(P(B\ | A') = \frac{\text{Number of outcomes}(B\cap A')}{\text{Number of outcomes}(A')} = \frac{16}{26}\) | M1, M1 |
| A1 | Final answer \(\frac{16}{26}, \frac{8}{13}, 0.6153846\) to at least 3SF |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(A) = \frac{10}{36}$, $P(B) = \frac{24}{36}$ | B1 | Accept $P(A) = \frac{10}{36}, \frac{5}{18}, 0.278$ and $P(B) = \frac{24}{36}, \frac{2}{3}, 0.667$ |
| $P(A \cap B) = \frac{8}{36}$ | B1 | |
| $\frac{10}{36} \times \frac{24}{36}$ | M1 | Their $P(A) \times$ their $P(B)$ seen numerically, $0 \leq$ their $P(A), P(B) \leq 1$ |
| $= \frac{5}{27}, 0.185 \left[\neq \frac{8}{36}\right]$; Events are not independent | A1 FT | Multiplication evaluated correctly and compared with intersection that is not a product of multiplication, conclusion stated, notation $P(A)$, $P(B)$ and $P(A\cap B)$ used |
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## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P(B\|A') = \frac{P(B\cap A')}{P(A')} = \right] \frac{\frac{16}{36}}{\left(1-\frac{10}{36}\right)}$ | M1 | $[P(B\cap A') =] \frac{16}{36}, 0.4444$ or their $P(B) -$ their $P(A\cap B)$ seen as numerator or denominator of conditional probability fraction |
| | M1 | $[P(A') =]\left(1-\frac{10}{36}\right), \frac{26}{36}, 0.7222$ or $1-$ their $P(A)$ seen as denominator of conditional probability fraction |
| $= \frac{8}{13}$ | A1 | Final answer $\frac{16}{26}, \frac{8}{13}, 0.6153846$ to at least 3SF |
| **Alternative (direct from outcome tables):** $P(B\|A') = \frac{\text{Number of outcomes}(B\cap A')}{\text{Number of outcomes}(A')} = \frac{16}{26}$ | M1, M1 | 16 seen as numerator; 26 seen as denominator |
| | A1 | Final answer $\frac{16}{26}, \frac{8}{13}, 0.6153846$ to at least 3SF |5 Jasmine throws two ordinary fair 6-sided dice at the same time and notes the numbers on the uppermost faces. The events $A$ and $B$ are defined as follows.\\
$A$ : The sum of the two numbers is less than 6 .\\
$B : \quad$ The difference between the two numbers is at most 2 .
\begin{enumerate}[label=(\alph*)]
\item Determine whether or not the events $A$ and $B$ are independent.
\item Find $\mathrm { P } \left( B \mid A ^ { \prime } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q5 [7]}}