Section B: Probability and Statistics [60 marks]

1. Express $\text{P}(A)$ and $\text{P}(B)$ in terms of $\text{P}(A \cap B)$ using the conditional probability formulas.

$\text{P}(B \mid A) = \frac{\text{P}(A \cap B)}{\text{P}(A)} \implies \text{P}(A) = \frac{\text{P}(A \cap B)}{0.2} = 5\text{P}(A \cap B)$
$\text{P}(A \mid B) = \frac{\text{P}(A \cap B)}{\text{P}(B)} \implies \text{P}(B) = \frac{\text{P}(A \cap B)}{0.6} = \frac{5}{3}\text{P}(A \cap B)$

2. Use the complement rule and De Morgan's laws to evaluate $\text{P}(A \cup B)$.

$\text{P}(A' \cap B') = 1 - \text{P}(A \cup B)$
$0.3 = 1 - \text{P}(A \cup B) \implies \text{P}(A \cup B) = 0.7$

3. Use the addition rule to form an equation and solve for $\text{P}(A \cap B)$.

$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) - \text{P}(A \cap B)$
$0.7 = 5\text{P}(A \cap B) + \frac{5}{3}\text{P}(A \cap B) - \text{P}(A \cap B)$
$0.7 = \left( 4 + \frac{5}{3} \right) \text{P}(A \cap B)$
$0.7 = \frac{17}{3} \text{P}(A \cap B)$
$\text{P}(A \cap B) = \frac{0.7 \times 3}{17} = \frac{2.1}{17} = \frac{21}{170}$

1. Find $\text{P}(Y)$ using the given intersections.

The event $Y$ can be partitioned into two mutually exclusive parts, $X \cap Y$ and $X' \cap Y$:
$\text{P}(Y) = \text{P}(X \cap Y) + \text{P}(X' \cap Y)$
$\text{P}(Y) = 0.1 + 0.35 = 0.45$

2. Express $\text{P}(X \cup Y \cup Z)$ using the addition rule to form an equation linking $\text{P}(X)$ and $\text{P}(Z)$.

Using the addition rule on the events $(X \cup Z)$ and $Y$:
$\text{P}(X \cup Y \cup Z) = \text{P}(X \cup Z) + \text{P}(Y) - \text{P}((X \cup Z) \cap Y)$
Since $X$ and $Z$ are mutually exclusive ($\text{P}(X \cap Z) = 0$), we have $\text{P}(X \cup Z) = \text{P}(X) + \text{P}(Z)$.
Furthermore, the intersection with $Y$ gives $(X \cup Z) \cap Y = (X \cap Y) \cup (Z \cap Y)$. Because $X \cap Z = \emptyset$, these two components are also mutually exclusive:
$\text{P}((X \cap Y) \cup (Z \cap Y)) = \text{P}(X \cap Y) + \text{P}(Y \cap Z)$
$= 0.1 + 0.2 = 0.3$
Substitute these into the addition rule equation:
$0.95 = (\text{P}(X) + \text{P}(Z)) + 0.45 - 0.3$
$0.95 = \text{P}(X) + \text{P}(Z) + 0.15$
$\text{P}(X) + \text{P}(Z) = 0.8$

3. Establish bounds for $\text{P}(X)$ based on the minimum possible probabilities of subsets.

Since a subset's probability cannot exceed the whole event's probability (and probabilities must be non-negative):
$X \cap Y \subseteq X \implies \text{P}(X) \ge \text{P}(X \cap Y) = 0.1$
$Y \cap Z \subseteq Z \implies \text{P}(Z) \ge \text{P}(Y \cap Z) = 0.2$

Substitute $\text{P}(Z) = 0.8 - \text{P}(X)$ into the second inequality:
$0.8 - \text{P}(X) \ge 0.2$
$\text{P}(X) \le 0.6$

Combining the bounds, we have:
$0.1 \le \text{P}(X) \le 0.6$