| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Combined event algebra |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability formulas (addition rule, complement rule, conditional probability, and independence test) with no conceptual challenges. All parts follow directly from standard S1 techniques with simple arithmetic, making it easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Answer | Marks |
|---|---|
| \(0.76 = 0.5 + 0.42 - P(A \cap B)\) | M1 |
| \(P(A \cap B) = 0.92 - 0.76 = 0.16\) | M1 A1 |
| Answer | Marks |
|---|---|
| \((1 - 0.5) + 0.16 = 0.66\) | M2 A1 |
| Answer | Marks |
|---|---|
| \(\frac{P(B \cap A')}{P(A')} = \frac{0.42 - 0.16}{1 - 0.5} = 0.52\) | M2 A1 |
| Answer | Marks |
|---|---|
| \(P(A) \times P(B) = 0.5 \times 0.42 = 0.21\) | M1 A1 |
| \(\neq P(A \cap B) \therefore\) not independent | A1 |
| (12) |
**Part (a)**
$0.76 = 0.5 + 0.42 - P(A \cap B)$ | M1 |
$P(A \cap B) = 0.92 - 0.76 = 0.16$ | M1 A1 |
**Part (b)**
$(1 - 0.5) + 0.16 = 0.66$ | M2 A1 |
**Part (c)**
$\frac{P(B \cap A')}{P(A')} = \frac{0.42 - 0.16}{1 - 0.5} = 0.52$ | M2 A1 |
**Part (d)**
$P(A) \times P(B) = 0.5 \times 0.42 = 0.21$ | M1 A1 |
$\neq P(A \cap B) \therefore$ not independent | A1 |
| (12) |4. The events $A$ and $B$ are such that
$$\mathrm { P } ( A ) = 0.5 , \mathrm { P } ( B ) = 0.42 \text { and } \mathrm { P } ( A \cup B ) = 0.76$$
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( A \cap B )$,
\item $\quad \mathrm { P } \left( A ^ { \prime } \cup B \right)$,
\item $\mathrm { P } \left( B \mid A ^ { \prime } \right)$.
\item Show that events $A$ and $B$ are not independent.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q4 [12]}}