# Question 6:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $A, C$ or $A, D$ or $B, D$ [Allow things like $A \cap D$] | B1 | For a correct pair; if more than one pair given then all must be correct; $P(A)$ and $P(C)$ etc is B0; $P(A \cap C) = 0$ is B0 but condone $A \cap C = \varnothing$ |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(C) = 0.6$ and $P(B) = p + 0.32$ and $P(B \cap C) = 0.27$; or $(0.08 + 0.25 + 0.27) \times (0.27 + 0.05 + p) = 0.27$ or $0.27 + 0.05 + p = \dfrac{0.27}{0.6} = 0.45$ | M1 | For all relevant labelled probabilities listed or a correct equation/expression for $p$ |
| $[p + 0.32 = 0.45$ so$]\ p = \mathbf{0.13}$ | A1 | |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P(A \mid B')\right] = \dfrac{q}{q+r+0.25+0.08}$ or $\dfrac{q}{1-(0.05+\text{"0.13"}+0.27)}$ or $\dfrac{q}{0.55}$ | M1 | For a correct method for $P(A \mid B')$ in $q$ (and $r$); ft their $p$; may be done in stages |
| $q + r = 1 - 0.65 - \text{"0.13"}\ [= 0.22]$ | M1 | For a correct equation for $q + r$ (o.e.)(ft their $p$); can accept $r = 0$ and $q = 0.22$ |
| Since $r \ldots 0$ the greatest value of $q$ is "0.22" so $P(A \mid B')\ ,,\ \mathbf{0.4}$ or $\dfrac{2}{5}$ | A1 | For 0.4 i.e. deducing the maximum value of $P(A \mid B')$; allow ",," 0.4 or $P(A \mid B') = 0.4$ |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P(B \mid A')\right] = \dfrac{0.27 + \text{"0.13"}}{0.6 + \text{"0.13"} + r} = 0.5$ or $\dfrac{0.27 + \text{"0.13"}}{1-(q+0.05)} = 0.5$ | M1 | For a correct equation for $r$ (or $q$) only; can have $p$ or ft their value for $p$; may be in stages |
| $r = \mathbf{0.07}$ | A1 | |
| $q = \mathbf{0.15}$ | A1ft | |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P\left([A \cup B]' \cap C\right)\right] = [0.25 + 0.08] = \mathbf{0.33}$ | B1 | For 0.33 |

## Part (f)
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $B \cap [A \cup C]'$ or $B \cap A' \cap C'$ or $(B \cap A') \cap (B \cap C')$ o.e. | B1 | For any correct expression; do **not** condone $P(\ldots$ |

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