| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Complement and union/intersection laws |
| Difficulty | Standard +0.8 This S4 question requires systematic application of conditional probability definitions and careful Venn diagram reasoning across three events. Part (i) involves non-trivial manipulation of P(A|B') to find intersections, parts (ii) test understanding of independence/mutual exclusivity, and part (iii) requires setting up and solving equations using the inclusion-exclusion principle with multiple constraints—more demanding than typical A-level probability but standard for S4. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(A \cap B') = 0.75 \times 0.4 = 0.3\) | M1A1 | |
| \(P(A \cap B) = 0.5 - \text{"0.3"} = 0.2\) | M1A1 | |
| \(P(A \cup B) = 0.5 + 0.6 - \text{"0.2"} = 0.9\) | M1A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No, \(P(A \cap B) \neq 0\) oe | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No, \(0.5 \times 0.6 \neq 0.2\) oe | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(A' \cap B' \cap C) = 0.1\) soi | B1ft | \(1-(i)\) |
| \(x = 0.1\) | B1 | |
| \(P(C) = 2x + 0.05 + 0.1 = 0.35\) | B1 | |
| [3] |
# Question 5:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A \cap B') = 0.75 \times 0.4 = 0.3$ | M1A1 | |
| $P(A \cap B) = 0.5 - \text{"0.3"} = 0.2$ | M1A1 | |
| $P(A \cup B) = 0.5 + 0.6 - \text{"0.2"} = 0.9$ | M1A1 | |
| **[6]** | | |
## Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No, $P(A \cap B) \neq 0$ oe | B1 | |
| **[1]** | | |
## Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| No, $0.5 \times 0.6 \neq 0.2$ oe | B1 | |
| **[1]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(A' \cap B' \cap C) = 0.1$ soi | B1ft | $1-(i)$ |
| $x = 0.1$ | B1 | |
| $P(C) = 2x + 0.05 + 0.1 = 0.35$ | B1 | |
| **[3]** | | |
---5 Events $A$ and $B$ are such that $\mathrm { P } ( A ) = 0.5 , \mathrm { P } ( B ) = 0.6$ and $\mathrm { P } \left( A \mid B ^ { \prime } \right) = 0.75$.\\
(i) Find $\mathrm { P } ( A \cap B )$ and $\mathrm { P } ( A \cup B )$.\\
(ii) Determine, giving a reason in each case,
\begin{enumerate}[label=(\alph*)]
\item whether $A$ and $B$ are mutually exclusive,
\item whether $A$ and $B$ are independent.\\
(iii) A further event $C$ is such that $\mathrm { P } ( A \cup B \cup C ) = 1$ and $\mathrm { P } ( A \cap B \cap C ) = 0.05$. It is also given that $\mathrm { P } \left( A \cap B ^ { \prime } \cap C \right) = \mathrm { P } \left( A ^ { \prime } \cap B \cap C \right) = x$ and $\mathrm { P } \left( A \cap B ^ { \prime } \cap C ^ { \prime } \right) = 2 x$.\\
Find $\mathrm { P } ( C )$.
\end{enumerate}
\hfill \mbox{\textit{OCR S4 2016 Q5 [11]}}