S1 January 2013 Q7
7. Given that
\[\mathrm{P}(A) = 0.35, \quad \mathrm{P}(B) = 0.45 \quad \text{and} \quad \mathrm{P}(A \cap B) = 0.13\]find
The event \(C\) has \(\mathrm{P}(C) = 0.20\)
The events \(A\) and \(C\) are mutually exclusive and the events \(B\) and \(C\) are independent.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(A \cup B) = 0.35 + 0.45 - 0.13\) or \(0.22 + 0.13 + 0.32\) | M1 |
| \(= \underline{\mathbf{0.67}}\) | A1 |
| (2) |
Notes
NB May see Venn diagram for \(A\) and \(B\) only used for (a) and (b) but M marks are awarded for correct expressions only. No ft from an incorrect diagram for M marks.
M1 for attempt to use the addition rule. Correct substitution i.e. correct expression seen
A1 for 0.67 only. Correct answer only scores 2/2
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(A^{\prime} \mid B^{\prime}) = \dfrac{\mathrm{P}(A^{\prime} \cap B^{\prime})}{\mathrm{P}(B^{\prime})}\) or \(\dfrac{0.33}{0.55}\) | M1 |
| \(= \dfrac{3}{5}\) or 0.6 | A1 |
| (2) |
Notes
M1 for a correct ratio of probabilities or a correct formula and at least one correct prob
For a correct formula allow “1 − their (a)” instead of 0.33 but not for correct ratio case.
Do not award for assuming independence i.e. \(\frac{\mathrm{P}(A^{\prime} \cap B^{\prime})}{\mathrm{P}(B^{\prime})} = \frac{0.65\times 0.55}{0.55}\) is M0. M0 if num>denom
A1 for 3/5 or any exact equivalent.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(B \cap C) = 0.45\times 0.2\) | M1 |
| \(= \underline{\mathbf{0.09}}\) | A1 |
| (2) |
Notes
M1 for correct expression. Need correct values for \(\mathrm{P}(B)\) and \(\mathrm{P}(C)\) seen.
A1 for 0.09 or any exact equivalent. Correct answer only is 2/2
| Scheme | Marks |
|---|---|
 Do not accept “blank” for zero | B1 B1ft B1 B1 |
| (4) |
Notes
No labels \(A\), \(B\), \(C\) in (d) loses 1st B1 but can score the other 3 by implication
B1 for box with \(B\) intersecting \(A\) and \(C\) but \(C\) not intersecting \(A\). No box is B0
B1ft for 0.13 and their 0.09 in correct places. [ft \(\mathrm{P}(B \cap C)\) from (c)]
B1 for any 2 of 0.22, 0.22, 0.11 and 0.23 correct
B1 for all 4 values correct
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(B \cup C)^{\prime} = 0.22 + \underline{0.22}\) or \(1 - [0.56]\) or \(1 - [0.13 + 0.23 + 0.09 + 0.11]\) o.e. | M1 |
| \(= \underline{\mathbf{0.44}}\) | A1 |
| (2) | |
| (12 marks) |
Notes
M1 for a correct expression or follow through from their Venn diagram
NB \(\mathrm{P}(B^{\prime})\times\mathrm{P}(C^{\prime}) = 0.55\times 0.8\) is OK. Do not ft “blank” for zero and M0 for negative probs.
A1 for 0.44 only. Correct answer only is 2/2