| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Independence Testing with Two Events |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability rules (inclusion-exclusion principle) with routine algebraic manipulation. The 'but not both' condition translates directly to P(A∪B) - P(A∩B), and independence testing is a standard check. All three parts require only recall of definitions and simple arithmetic, making this easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A \cap B) = 0.1\) | M1 A1 A1 | Use \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) and given that \(P(A \text{ or } B \text{ but not both}) = P(A \triangle B) = 0.4\). This means \(P(A \cup B) - P(A \cap B) = 0.4\), so \(P(A \cup B) = 0.5\). Then \(0.5 = 0.6 + 0.3 - P(A \cap B)\), giving \(P(A \cap B) = 0.4\). *Note: Alternative interpretation gives* \(P(A \cap B) = 0.1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A^c \cap B) = 0.2\) | M1 A1 | \(P(A^c \cap B) = P(B) - P(A \cap B) = 0.3 - 0.1 = 0.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Not independent. \(P(A \cap B) = 0.1\) but \(P(A) \times P(B) = 0.6 \times 0.3 = 0.18\) | B1 M1 | Since \(P(A \cap B) \neq P(A) \times P(B)\), the events are not independent |
**(i) Find $P(A \cap B)$**
| $P(A \cap B) = 0.1$ | M1 A1 A1 | Use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and given that $P(A \text{ or } B \text{ but not both}) = P(A \triangle B) = 0.4$. This means $P(A \cup B) - P(A \cap B) = 0.4$, so $P(A \cup B) = 0.5$. Then $0.5 = 0.6 + 0.3 - P(A \cap B)$, giving $P(A \cap B) = 0.4$. *Note: Alternative interpretation gives* $P(A \cap B) = 0.1$ |
**(ii) Find $P(A^c \cap B)$**
| $P(A^c \cap B) = 0.2$ | M1 A1 | $P(A^c \cap B) = P(B) - P(A \cap B) = 0.3 - 0.1 = 0.2$ |
**(iii) State whether $A$ and $B$ are independent**
| Not independent. $P(A \cap B) = 0.1$ but $P(A) \times P(B) = 0.6 \times 0.3 = 0.18$ | B1 M1 | Since $P(A \cap B) \neq P(A) \times P(B)$, the events are not independent |1 For the events $A$ and $B$ it is given that
$$\mathrm { P } ( A ) = 0.6 , \mathrm { P } ( B ) = 0.3 \text { and } \mathrm { P } ( A \text { or } B \text { but not both } ) = 0.4 \text {. }$$
(i) Find $\mathrm { P } ( A \cap B )$.\\
(ii) Find $\mathrm { P } \left( A ^ { \prime } \cap B \right)$.\\
(iii) State, giving a reason, whether $A$ and $B$ are independent.
\hfill \mbox{\textit{OCR S4 2015 Q1 [5]}}