| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question requiring application of standard probability laws (complement rule, addition rule) and basic algebraic manipulation. While it involves multiple steps and working with an algebraic parameter, the techniques are routine for S4 level and the deduction about the range of c follows directly from probability constraints (0 ≤ P(B|A) ≤ 1). Slightly above average difficulty due to the algebraic element and multi-part reasoning, but still a standard textbook-style exercise. |
| 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 | Mark | Guidance |
| Table with row/column headings copied (A:500,400,700,600; B:300,200,400,350; C:500,300,750,680) | B1 | For copying the table, with row and column headings (accept consistent scalings) |
| Dummy row \(D\) with all zeros | B1 | For dummy row (Daniel) with all equal values [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Reduce rows: 100,0,300,200 / 100,0,200,150 / 200,0,450,380 / 0,0,0,0 | M1 | For a substantially correct attempt at reducing rows and columns |
| Correct reduced cost matrix | A1 | For correct reduced cost matrix (ft scalings); do not treat as MR |
| Columns are already reduced | [2] | |
| Cover zeros using two lines | M1 | For covering zeros using minimum number of lines, clearly seen or implied from augmenting |
| Single augmentation by 100 | M1 dep | For a single augmentation by 100 (ft their matrix); accept either way of augmenting by 100 |
| Augmented matrix: 0,0,200,100 / 0,0,100,50 / 100,0,350,280 / 0,100,0,0 | A1 ft | For a correct augmented matrix (ft their matrix) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Cover zeros using three lines | M1 | For covering zeros using minimum number of lines a second time, clearly seen or implied from augmenting |
| Augment by 50 | M1 dep | For a single augmentation by 50 (ft their matrix); accept either way of augmenting by 50 |
| Matrix: 0,0,150,50 / 0,0,50,0 / 100,0,300,230 / 50,150,0,0 | A1 ft | For a correct augmented matrix (ft their matrix) |
| Complete matching achieved | B1 | For a complete matching achieved; must follow from an attempt at reducing or augmenting a matrix, not just implied from a list of the matching [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Allclean \(\to\) house 1; Brightenupp \(\to\) house 4; Clean4U \(\to\) house 2 | B1 | For \(A=1,\ B=4,\ C=2\) (may also list \(D=3\)) cao |
| Cost \(= £1150\) | B1 | For 1150 cao [2] |
# Question 1 (Paper 4737):
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table with row/column headings copied (A:500,400,700,600; B:300,200,400,350; C:500,300,750,680) | B1 | For copying the table, with row and column headings (accept consistent scalings) |
| Dummy row $D$ with all zeros | B1 | For dummy row (Daniel) with all equal values **[2]** |
## Part (ii) - first augmentation:
| Answer | Mark | Guidance |
|--------|------|----------|
| Reduce rows: 100,0,300,200 / 100,0,200,150 / 200,0,450,380 / 0,0,0,0 | M1 | For a substantially correct attempt at reducing rows and columns |
| Correct reduced cost matrix | A1 | For correct reduced cost matrix (ft scalings); do not treat as MR |
| Columns are already reduced | | **[2]** |
| Cover zeros using two lines | M1 | For covering zeros using minimum number of lines, clearly seen or implied from augmenting |
| Single augmentation by 100 | M1 dep | For a single augmentation by 100 (ft their matrix); accept either way of augmenting by 100 |
| Augmented matrix: 0,0,200,100 / 0,0,100,50 / 100,0,350,280 / 0,100,0,0 | A1 ft | For a correct augmented matrix (ft their matrix) **[3]** |
## Part (ii) - second augmentation:
| Answer | Mark | Guidance |
|--------|------|----------|
| Cover zeros using three lines | M1 | For covering zeros using minimum number of lines a second time, clearly seen or implied from augmenting |
| Augment by 50 | M1 dep | For a single augmentation by 50 (ft their matrix); accept either way of augmenting by 50 |
| Matrix: 0,0,150,50 / 0,0,50,0 / 100,0,300,230 / 50,150,0,0 | A1 ft | For a correct augmented matrix (ft their matrix) |
| Complete matching achieved | B1 | For a complete matching achieved; must follow from an attempt at reducing or augmenting a matrix, not just implied from a list of the matching **[4]** |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Allclean $\to$ house 1; Brightenupp $\to$ house 4; Clean4U $\to$ house 2 | B1 | For $A=1,\ B=4,\ C=2$ (may also list $D=3$) cao |
| Cost $= £1150$ | B1 | For 1150 cao **[2]** |
---1 For the events $A$ and $B , \mathrm { P } ( A ) = 0.3 , \mathrm { P } ( B ) = 0.6$ and $\mathrm { P } \left( A ^ { \prime } \cap B ^ { \prime } \right) = c$, where $c \neq 0$.\\
(i) Find $\mathrm { P } ( A \cap B )$ in terms of $c$.\\
(ii) Find $\mathrm { P } ( B \mid A )$ and deduce that $0.1 \leqslant c \leqslant 0.4$.
\hfill \mbox{\textit{OCR S4 2007 Q1 [6]}}