事件 A、B 和 C 满足 P(A) = 0.8，P(B) = 0.2 且 P(C) = 0.6。已知 $P(A \cap B) = P(B)$ 且 $P(B \cap C) = 0.1$。

1. Determine the probability of $A \cap B \cap C$ using the given conditions.

Given $\mathrm{P}(A \cap B) = \mathrm{P}(B) = 0.2$, all outcomes of $B$ must lie entirely within $A$ (i.e. $B \subseteq A$). This means $\mathrm{P}(A^c \cap B) = 0$.
Since $B$ is a subset of $A$, the intersection $B \cap C$ is identical to $A \cap B \cap C$.
Therefore, $\mathrm{P}(A \cap B \cap C) = \mathrm{P}(B \cap C) = 0.1$.

2. Form an inequality to find the maximum possible value by partitioning $C$.

Let $x = \mathrm{P}(A \cap B^c \cap C)$.
We can partition the set $C$ into four mutually exclusive regions:
$\mathrm{P}(C) = \mathrm{P}(A \cap B \cap C) + \mathrm{P}(A \cap B^c \cap C) + \mathrm{P}(A^c \cap B \cap C) + \mathrm{P}(A^c \cap B^c \cap C)$
$0.6 = 0.1 + x + 0 + \mathrm{P}(A^c \cap B^c \cap C)$
$\mathrm{P}(A^c \cap B^c \cap C) = 0.5 - x$
Since all probabilities must be non-negative, we require:
$0.5 - x \ge 0 \implies x \le 0.5$
Thus, the maximum possible value is $0.5$.

3. Form an inequality to find the minimum possible value using the addition rule.

The total probability of the union of $A$ and $C$ cannot exceed $1$, so $\mathrm{P}(A \cup C) \le 1$.
Using the addition rule for probabilities:
$\mathrm{P}(A \cup C) = \mathrm{P}(A) + \mathrm{P}(C) - \mathrm{P}(A \cap C)$
We can express $\mathrm{P}(A \cap C)$ as the sum of its mutually exclusive parts:
$\mathrm{P}(A \cap C) = \mathrm{P}(A \cap B \cap C) + \mathrm{P}(A \cap B^c \cap C) = 0.1 + x$
Substituting this into the union inequality:
$0.8 + 0.6 - (0.1 + x) \le 1$
$1.4 - 0.1 - x \le 1$
$1.3 - x \le 1 \implies x \ge 0.3$
Thus, the minimum possible value is $0.3$.

1. Determine the valid range for $\text{P}(A \cap B^c \cap C)$ (required context from part a(i)).

Since $\text{P}(A \cap B) = \text{P}(B) = 0.2$, event $B$ is a subset of $A$. This implies $\text{P}(A \cap B \cap C) = \text{P}(B \cap C) = 0.1$.
Let $y = \text{P}(A \cap B^c \cap C)$. The intersection $A \cap C$ is partitioned into $A \cap B \cap C$ and $A \cap B^c \cap C$, so $\text{P}(A \cap C) = 0.1 + y$.
Using the addition rule, $\text{P}(A \cup C) = \text{P}(A) + \text{P}(C) - \text{P}(A \cap C) = 0.8 + 0.6 - (0.1 + y) = 1.3 - y$.
To find the valid range for $y$, we consider boundary probabilities:

• $\text{P}(A \cup C) \le 1 \implies 1.3 - y \le 1 \implies y \ge 0.3$.
• $\text{P}(A \cap C) \le \text{P}(C) \implies 0.1 + y \le 0.6 \implies y \le 0.5$.
  Thus, $0.3 \le y \le 0.5$.

2. Apply the conditional probability formula and simplify the intersection.

By the conditional probability formula:
$\text{P}(A \cap B^c \cap C \mid A \cup C) = \frac{\text{P}((A \cap B^c \cap C) \cap (A \cup C))}{\text{P}(A \cup C)}$
Since $A \cap B^c \cap C$ is a subset of $A \cap C$, which is a subset of $A \cup C$, the intersection simplifies to just the event $A \cap B^c \cap C$:
$\text{P}(A \cap B^c \cap C \mid A \cup C) = \frac{\text{P}(A \cap B^c \cap C)}{\text{P}(A \cup C)}$

3. Express the conditional probability strictly in terms of $y$.

Substitute the expressions derived in Step 1 into the simplified fraction:
$\text{P}(A \cap B^c \cap C \mid A \cup C) = \frac{y}{1.3 - y}$
Let $f(y) = \frac{y}{1.3 - y}$.

4. Determine the maximum and minimum values of the rational function over the domain of $y$.

To find the extreme values over $0.3 \le y \le 0.5$, rewrite $f(y)$:
$f(y) = \frac{-(1.3 - y) + 1.3}{1.3 - y} = \frac{1.3}{1.3 - y} - 1$
As $y$ increases, the denominator $(1.3 - y)$ decreases, causing the fraction to strictly increase.

• Minimum value (at $y=0.3$):
  $f(0.3) = \frac{0.3}{1.3 - 0.3} = 0.3$
• Maximum value (at $y=0.5$):
  $f(0.5) = \frac{0.5}{1.3 - 0.5} = \frac{0.5}{0.8} = \frac{5}{8}$

1. Calculate $P(A \cap C)$ using the independence of $A$ and $C$.

Since $A$ and $C$ are independent,
$P(A \cap C) = P(A) \times P(C)$
$P(A \cap C) = 0.8 \times 0.6 = 0.48$

2. Determine $P(A \cap B \cap C)$ using the given condition $P(A \cap B) = P(B)$.

We are given $P(A \cap B) = P(B) = 0.2$.
Since $P(A^c \cap B) = P(B) - P(A \cap B) = 0.2 - 0.2 = 0$, event $B$ only occurs within event $A$ (i.e. $B \subset A$).
Therefore, $A \cap B = B$, which means:
$P(A \cap B \cap C) = P(B \cap C) = 0.1$

3. Calculate $P(A \cap B^c \cap C)$ using the partition of $A \cap C$.

The intersection $A \cap C$ can be split into two mutually exclusive parts: those inside $B$ and those outside $B$.
$P(A \cap C) = P(A \cap B \cap C) + P(A \cap B^c \cap C)$
$0.48 = 0.1 + P(A \cap B^c \cap C)$
$P(A \cap B^c \cap C) = 0.48 - 0.1 = 0.38$