June 2024 Paper 3 Q6
6. The Venn diagram, where \(p\), \(q\) and \(r\) are probabilities, shows the events \(A\), \(B\), \(C\) and \(D\) and associated probabilities.
The events \(B\) and \(C\) are independent.
Given that \(\mathrm{P}(B \mid A^{\prime}) = 0.5\)
| Scheme | Marks | AO |
|---|---|---|
| \(A, C\) or \(A, D\) or \(B, D\) [Allow things like \(A \cap D\)] | B1 | 1.2 |
| (1) |
Notes
B1 for a correct pair. If more than one pair is given then all must be correct.
\(\mathrm{P}(A)\) and \(\mathrm{P}(C)\) etc is B0 \(\mathrm{P}(A \cap C) = 0\) is B0 but condone things like \(A \cap C = \varnothing\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(C) = 0.6\) and \(\mathrm{P}(B) = p + 0.32\) and \(\mathrm{P}(B \cap C) = 0.27\) or \((0.08 + 0.25 + 0.27) \times (0.27 + 0.05 + p) = 0.27\) or \(0.27 + 0.05 + p = \tfrac{0.27}{0.6} = 0.45\) | M1 | 1.1b |
| [\(p + 0.32 = 0.45\) so] \(p = \underline{\mathbf{0.13}}\) | A1 | 2.2a |
| (2) |
Notes
In parts (b) – (d) we will condone poor notation and mark equations/expressions
M1 for all relevant labelled probabilities listed or a correct equation/expression for \(p\)
A1 for \(p = 0.13\)
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{P}(A \mid B^{\prime})\right] = \dfrac{q}{q + r + 0.25 + 0.08}\) or \(\dfrac{q}{1 - (0.05 + \text{``}0.13\text{''} + 0.27)}\) or \(\dfrac{q}{0.55}\) | M1 | 2.1 |
| \(q + r = 1 - 0.65 - \text{“}0.13\text{”}\) [\(= 0.22\)] | M1 | 1.1b |
| Since \(r \geqslant 0\) the greatest value of \(q\) is “0.22” so \(\mathrm{P}(A \mid B^{\prime}) \leqslant \underline{\mathbf{0.4}}\) or \(\underline{\tfrac{2}{5}}\) | A1 | 2.2a |
| (3) |
Notes
In parts (b) – (d) we will condone poor notation and mark equations/expressions
In parts (c) and (d) they can use letter \(p\) or we ft their value for \(p\) provided a probability
1st M1 for a correct method for \(\mathrm{P}(A \mid B^{\prime})\) in \(q\) (and \(r\)) ft their \(p\). May be done in stages
e.g. find correct expression for \(\mathrm{P}(B^{\prime})\), simplify incorrectly then use \(q\) over this
2nd M1 for a correct equation for \(q + r\) (o.e.)(ft their \(p\)) Can accept \(r = 0\) and \(q = 0.22\)
NB sight of \(\tfrac{0.22}{0.55}\) will score M1M1
A1 for 0.4 i.e. deducing the maximum value of \(\mathrm{P}(A \mid B^{\prime})\). Allow \(\leqslant 0.4\) or \(\mathrm{P}(A \mid B^{\prime}) = 0.4\)
Can award 3/3 for \(\mathrm{P}(A \mid B^{\prime}) = 0.4\) but not 0.4 alone as it can come from e.g \(\mathrm{P}(C^{\prime})\)
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{P}(B \mid A^{\prime}) =\right] \dfrac{0.27 + \text{``}0.13\text{''}}{0.6 + \text{``}0.13\text{''} + r} = 0.5\) or \(\dfrac{0.27 + \text{``}0.13\text{''}}{1 - (q + 0.05)} = 0.5\) | M1 | 1.1b |
| \(r = \underline{\mathbf{0.07}}\), | A1 | 1.1b |
| \(q = \underline{\mathbf{0.15}}\) | A1ft | 1.1b |
| (3) |
Notes
In parts (b) – (d) we will condone poor notation and mark equations/expressions
In parts (c) and (d) they can use letter \(p\) or we ft their value for \(p\) provided a probability
M1 for a correct equation for \(r\) (or \(q\)) only can have \(p\) or ft their value for \(p\).
May be in stages
e.g. find \(\mathrm{P}(A^{\prime}) = 0.27 + 0.25 + 0.08 + p + r\) but make a slip in getting 0.6 then use this.
1st A1 for \(r = 0.07\) or \(q = 0.15\)
2nd A1ft for \(r = 0.07\) and \(q = 0.15\) or values giving \(q + r = 0.22\) provided both \(q\) and \(r\) are probabilities. Obviously, 2nd A1ft is dependent on the M1
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\mathrm{P}\left(\left[A \cup B\right]^{\prime} \cap C\right) =\right]\left[0.25 + 0.08\right] = \underline{\mathbf{0.33}}\) | B1 | 1.1b |
| (1) |
Notes
B1 for 0.33
| Scheme | Marks | AO |
|---|---|---|
| e.g. \(B \cap \left[A \cup C\right]^{\prime}\) or \(B \cap A^{\prime} \cap C^{\prime}\) or \((B \cap A^{\prime}) \cap (B \cap C^{\prime})\) o.e. | B1 | 1.1b |
| (1) | ||
| (11 marks) |
Notes
B1 for any correct expression. Do not condone P(…