| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.8 This S4 question requires understanding of conditional probability definitions and algebraic manipulation with inequalities. Part (i) needs insight to use P(A∩B) ≤ P(A) combined with the conditional probability formula, while part (ii) involves rearranging P(A∪B) = P(A) + P(B) - P(A∩B) with P(A∩B) = P(A|B)P(B). More conceptually demanding than routine probability calculations but still within standard S4 scope. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(A \cap B)=0.6P(B)\) | M1 | May be implied. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B) \leq 0.3/0.6 = 0.5\) AG | M1, A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B) = (x – 0.3)/0.4\) | M1, A1 | Use formulae for union and cond prob. |
| [2] |
**(i)** $P(A \cap B)=0.6P(B)$ | M1 | May be implied.
$P(A \cap B) \leq P(A) = 0.3$
$P(B) \leq 0.3/0.6 = 0.5$ AG | M1, A1 | | 0.3P(B A)=0.6P(B) M1 ;use P(B A)⊆1 M1
| [3] |
**(ii)** $P(A \cup B) = x = 0.3 + P(B) – 0.6P(B)$
$P(B) = (x – 0.3)/0.4$ | M1, A1 | Use formulae for union and cond prob.
| [2] |8 Events $A$ and $B$ are such that $\mathrm { P } ( A ) = 0.3$ and $\mathrm { P } ( A \mid B ) = 0.6$.\\
(i) Show that $\mathrm { P } ( B ) \leqslant 0.5$.\\
(ii) Given also that $\mathrm { P } ( A \cup B ) = x$, find $\mathrm { P } ( B )$ in terms of $x$.
\hfill \mbox{\textit{OCR S4 2012 Q8 [5]}}