| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Moderate -0.8 This is a straightforward application of basic probability formulas: P(A∩B) = P(B|A)×P(A) for part (i), then using P(A∪B) = P(A) + P(B) - P(A∩B) to find P(B) in part (ii). Requires only direct substitution into standard formulas with no problem-solving insight needed, making it easier than average but not trivial since students must recall and correctly apply the conditional probability definition. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
**Question 2**
**(i)** $P(A \cap B) = P(A) \times P(B \mid A)$ — M1
$= \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ — A1 **[2]**
Conditional probability rule applied, s.o.i. c.a.o. Accept solutions based on Venn diagrams.
**(ii)** $P(B) = P(A \cup B) - P(A) + P(A \cap B)$ — M1
$= \frac{5}{6} - \frac{1}{2} + \frac{1}{8} = \frac{11}{24}$ — A1 **[2]**
Probability rule applied, s.o.i.
Ft (i) provided both $P(A \cap B)$ and $P(B)$ lie between 0 and 1.
**Total: [4]**2 Events $A$ and $B$ are such that $\mathrm { P } ( A ) = \frac { 1 } { 2 } , \mathrm { P } ( A \cup B ) = \frac { 5 } { 6 }$ and $\mathrm { P } ( B \mid A ) = \frac { 1 } { 4 }$.\\
Find\\
(i) $\mathrm { P } ( A \cap B )$,\\
(ii) $\mathrm { P } ( B )$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q2 [4]}}