对于事件 $A$ 和 $B$，已知 $\mathrm{P}(A) = 0.55$，$\mathrm{P}(B) = 0.65$ 且 $\mathrm{P}(A \cup B) = 0.93$。

1. State the addition rule for probabilities

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

2. Substitute the given values to find $\mathrm{P}(A \cap B)$

$0.93 = 0.55 + 0.65 - \mathrm{P}(A \cap B)$
$\mathrm{P}(A \cap B) = 1.20 - 0.93$
$\mathrm{P}(A \cap B) = 0.27$

1. Find the probability of the region outside both $A$ and $B$.

$\mathrm{P}((A \cup B)') = 1 - \mathrm{P}(A \cup B) = 1 - 0.93 = 0.07$

2. Use Venn diagram logic to express $A \cup B'$ as the sum of mutually exclusive components and calculate its value.

By visualising a Venn diagram, the region $A \cup B'$ consists of the circle $A$ and the region outside both circles $(A \cup B)'$.
Since $A$ and $(A \cup B)'$ are mutually exclusive:
$\mathrm{P}(A \cup B') = \mathrm{P}(A) + \mathrm{P}((A \cup B)') = 0.55 + 0.07 = 0.62$

1. Retrieve or calculate $\mathrm{P}(A \cap B)$ from part (i) and calculate $\mathrm{P}(A) \times \mathrm{P}(B)$

From part (i), we have $\mathrm{P}(A \cap B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cup B) = 0.55 + 0.65 - 0.93 = 0.27$.

Calculating the product of the individual probabilities:
$\mathrm{P}(A) \times \mathrm{P}(B) = 0.55 \times 0.65 = 0.3575$

2. Compare the values to determine independence

Since $0.27 \neq 0.3575$, we have $\mathrm{P}(A \cap B) \neq \mathrm{P}(A) \times \mathrm{P}(B)$.
Therefore, events $A$ and $B$ are not independent.

1. Determine the probabilities of regions available for event $C$.

From part (i), we have $\mathrm{P}(A \cap B) = 0.55 + 0.65 - 0.93 = 0.27$.

Since events $B$ and $C$ are mutually exclusive, $B \cap C = \emptyset$, which means $C$ must lie entirely within $B'$.
The region $B'$ can be partitioned into two parts: inside $A$ ($A \cap B'$) and outside $A$ ($A' \cap B'$).

$\mathrm{P}(A \cap B') = \mathrm{P}(A) - \mathrm{P}(A \cap B) = 0.55 - 0.27 = 0.28$
$\mathrm{P}(A' \cap B') = 1 - \mathrm{P}(A \cup B) = 1 - 0.93 = 0.07$

2. Find the greatest possible value of $\mathrm{P}(A \cap C)$.

For $\mathrm{P}(A \cap C)$ to be as large as possible, $C$ should lie inside $A \cap B'$ as much as possible.
Since $C$ cannot overlap with $B$, the event $A \cap C$ must be a subset of $A \cap B'$. Therefore:
$\mathrm{P}(A \cap C) \le \mathrm{P}(A \cap B') = 0.28$

Since $\mathrm{P}(C) = 0.3$, which is larger than $0.28$, it is possible for $C$ to completely cover $A \cap B'$.
Greatest possible value $= 0.28$

3. Find the least possible value of $\mathrm{P}(A \cap C)$.

For $\mathrm{P}(A \cap C)$ to be as small as possible, $C$ should lie completely outside $A$ to the maximum extent possible. The only available region for $C$ outside $A$ is $A' \cap B'$.

$\mathrm{P}(C \cap A') \le \mathrm{P}(A' \cap B') = 0.07$

Using the fact that $\mathrm{P}(C) = \mathrm{P}(C \cap A) + \mathrm{P}(C \cap A')$:
$\mathrm{P}(A \cap C) = \mathrm{P}(C) - \mathrm{P}(C \cap A')$
$\mathrm{P}(A \cap C) \ge 0.30 - 0.07 = 0.23$

Least possible value $= 0.23$