Section B: Statistics [60 marks]

1. Express the probability of $A' \cap B$ in terms of $B$ and $A \cap B$.

$P(A' \cap B) = P(B) - P(A \cap B)$

2. Apply the definition of independent events for $A$ and $B$, and factorise the expression.

Since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$.
$P(A' \cap B) = P(B) - P(A)P(B)$
$P(A' \cap B) = P(B)[1 - P(A)]$

3. Use the complement rule to complete the proof.

$P(A' \cap B) = P(B)P(A')$
$P(A' \cap B) = P(A')P(B)$
Since this product condition is satisfied, $A'$ and $B$ are also independent.

1. Express the probability of region $B$ only in terms of $b$.

$\text{Since } A \text{ and } B \text{ are independent, } P(A \cap B) = P(A)P(B) = 0.3b.$
$\text{Since } A \text{ and } C \text{ are mutually exclusive, } P(A \cap C) = 0.$
$\text{The probability of the region 'only } B\text{' in the Venn diagram is:}$
$P(B \cap A' \cap C') = P(B) - P(A \cap B) - P(B \cap C)$
$= b - 0.3b - 0.15 = 0.7b - 0.15$

2. Find the lower bound for $b$ by setting $P(B \text{ only}) \ge 0$.

$\text{Since all probabilities must be non-negative, } P(B \cap A' \cap C') \ge 0:$
$0.7b - 0.15 \ge 0 \implies 0.7b \ge 0.15 \implies b \ge \frac{0.15}{0.7} = \frac{15}{70} = \frac{3}{14}$

3. Use the sum of all Venn diagram regions to find the upper bound for $b$.

$\text{The sum of probabilities for all mutually exclusive regions in the Venn diagram is } 1:$
$P(A \text{ only}) + P(A \cap B) + P(B \text{ only}) + P(B \cap C) + P(C \text{ only}) + P(A' \cap B' \cap C') = 1$
$(0.3 - 0.3b) + 0.3b + (0.7b - 0.15) + 0.15 + P(C \text{ only}) + 0.2 = 1$
$0.5 + 0.7b + P(C \text{ only}) = 1 \implies P(C \text{ only}) = 0.5 - 0.7b$

$\text{Since } P(C \text{ only}) \ge 0:$
$0.5 - 0.7b \ge 0 \implies 0.7b \le 0.5 \implies b \le \frac{5}{7}$

4. Combine the bounds to find the range of values for $P(A \cap B)$.

$\text{We established that } \frac{3}{14} \le b \le \frac{5}{7}.$
$\text{Since } P(A \cap B) = 0.3b = \frac{3}{10}b, \text{ we multiply the inequality by } \frac{3}{10}:$
$\frac{3}{10} \times \frac{3}{14} \le P(A \cap B) \le \frac{3}{10} \times \frac{5}{7}$
$\frac{9}{140} \le P(A \cap B) \le \frac{3}{14}$