| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.8 This S4 conditional probability question requires systematic application of probability laws across three parts. Part (i) needs manipulation of conditional probability formula with complements. Part (iii) is particularly demanding, requiring construction of a Venn diagram with three events, using independence and conditional probability constraints simultaneously, plus the condition that A∪B∪C is the entire sample space—this involves solving a system with multiple algebraic relationships, which is beyond routine S4 exercises. |
| 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/Working | Marks | Guidance |
| \(P(A | B') = P(A \cap B') / P(B')\) | M1 |
| \(\Rightarrow P(A \cap B') = \frac{1}{8}\) AEF | A1 | May be implied |
| Use \(P(A \cap B) = P(A) - P(A \cap B')\) | M1 | |
| To give \(P(A \cap B) = \frac{5}{8}\) AEF | A1 4 | Or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A \cap B \cap C) = \frac{5}{8} \times \frac{1}{4} = \frac{5}{32}\) AEF | B1\(\checkmark\) 1 | Ft \(\frac{5}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B \cap C) = 3\lambda/4\) and \(P(C \cap A) = 3\lambda/4\) | M1 | For use of both conditional probs |
| Use formula for \(P(A \cup B \cup C)\) | M1 | Allow one sign error |
| And \(P(A \cup B \cup C) = 1\) | B1 | |
| Sub into formula for \(P(A \cup B \cup C)\) and solve for \(\lambda\) | M1 | |
| giving \(\lambda = \frac{3}{16}\) AEF | A1 5 | |
| [10] |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A|B') = P(A \cap B') / P(B')$ | M1 | |
| $\Rightarrow P(A \cap B') = \frac{1}{8}$ AEF | A1 | May be implied |
| Use $P(A \cap B) = P(A) - P(A \cap B')$ | M1 | |
| To give $P(A \cap B) = \frac{5}{8}$ AEF | A1 **4** | Or equivalent |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cap B \cap C) = \frac{5}{8} \times \frac{1}{4} = \frac{5}{32}$ AEF | B1$\checkmark$ **1** | Ft $\frac{5}{8}$ |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \cap C) = 3\lambda/4$ and $P(C \cap A) = 3\lambda/4$ | M1 | For use of both conditional probs |
| Use formula for $P(A \cup B \cup C)$ | M1 | Allow one sign error |
| And $P(A \cup B \cup C) = 1$ | B1 | |
| Sub into formula for $P(A \cup B \cup C)$ and solve for $\lambda$ | M1 | |
| giving $\lambda = \frac{3}{16}$ AEF | A1 **5** | |
| | **[10]** | |
---3 For the events $A$ and $B , \mathrm { P } ( A ) = \mathrm { P } ( B ) = \frac { 3 } { 4 }$ and $\mathrm { P } \left( A \mid B ^ { \prime } \right) = \frac { 1 } { 2 }$.\\
(i) Find $\mathrm { P } ( A \cap B )$.
For a third event $C , \mathrm { P } ( C ) = \frac { 1 } { 4 }$ and $C$ is independent of the event $A \cap B$.\\
(ii) Find $\mathrm { P } ( A \cap B \cap C )$.\\
(iii) Given that $\mathrm { P } ( C \mid A ) = \lambda$ and $\mathrm { P } ( B \mid C ) = 3 \lambda$, and that no event occurs outside $A \cup B \cup C$, find the value of $\lambda$.
\hfill \mbox{\textit{OCR S4 2011 Q3 [10]}}