| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Moderate -0.8 This is a straightforward probability question testing basic formulas: P(A∪B) = P(A) + P(B) - P(A∩B) for part (i), conditional probability P(B|A) = P(A∩B)/P(A) for part (ii), and checking independence via P(A∩B) = P(A)P(B) for part (iii). All parts involve direct substitution into standard formulas with no problem-solving or conceptual difficulty, making it easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3\) | M1 A1 | Probability rule applied, s.o.i. c.a.o. Accept solutions based on Venn diagrams. |
| (ii) \(P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.6} = 0.5\) | M1 A1 |
| (iii) \(A\) and \(B\) are independent since \(P(B | A) = P(B) = 0.5\) | B1 |
**(i)** $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3$ | M1 A1 | Probability rule applied, s.o.i. c.a.o. Accept solutions based on Venn diagrams.
**(ii)** $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3}{0.6} = 0.5$ | M1 A1 | Conditional probability rule applied, s.o.i. ft (i) provided both $P(A \cap B)$ and $P(B)$ lie between 0 and 1.
**(iii)** $A$ and $B$ are independent since $P(B|A) = P(B) = 0.5$ | B1 | ft (ii). Must be supported by explicit numerical evidence. Accept alternatives, e.g. $P(A \cap B) = P(A) \times P(B)$, with evidence.$A$ and $B$ are two events. You are given that $\mathrm{P}(A) = 0.6$, $\mathrm{P}(B) = 0.5$ and $\mathrm{P}(A \cup B) = 0.8$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm{P}(A \cap B)$. [2]
\item Find $\mathrm{P}(B | A)$. [2]
\item Explain whether the events $A$ and $B$ are independent or not. [1]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2014 Q2 [5]}}