已知事件 $A$、$B$ 和 $C$ 满足 $\mathrm{P}(A) = 0.1$、$\mathrm{P}(B) = 0.2$ 且 $\mathrm{P}(C) = c$。$A$ 和 $B$ 是独立事件。$A$ 和 $C$ 是互斥事件。

1. State the value of $\mathrm{P}(A | B)$ using the property of independent events

Since $A$ and $B$ are independent events, the occurrence of $B$ does not affect the probability of $A$.
$\mathrm{P}(A | B) = \mathrm{P}(A) = 0.1$

2. State the value of $\mathrm{P}(A | C)$ using the property of mutually exclusive events

Since $A$ and $C$ are mutually exclusive events, they cannot occur at the same time, meaning $\mathrm{P}(A \cap C) = 0$.
$\mathrm{P}(A | C) = \frac{\mathrm{P}(A \cap C)}{\mathrm{P}(C)} = 0$

1. Calculate $\mathrm{P}(A \cup B)$ using the independence of $A$ and $B$.

Since $A$ and $B$ are independent, $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) = 0.1 \times 0.2 = 0.02$.
$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$
$\mathrm{P}(A \cup B) = 0.1 + 0.2 - 0.02 = 0.28$

2. Express $\mathrm{P}(A' \cap C')$ in terms of $c$ using the fact that $A$ and $C$ are mutually exclusive.

Since $A$ and $C$ are mutually exclusive, $\mathrm{P}(A \cup C) = \mathrm{P}(A) + \mathrm{P}(C) = 0.1 + c$.
Using De Morgan's laws, $\mathrm{P}(A' \cap C') = 1 - \mathrm{P}(A \cup C)$.
$\mathrm{P}(A' \cap C') = 1 - (0.1 + c) = 0.9 - c$

3. Equate the two probabilities and solve for $c$.

Equating $\mathrm{P}(A \cup B)$ to $\mathrm{P}(A' \cap C')$:
$0.28 = 0.9 - c$
$c = 0.9 - 0.28 = 0.62$

1. Express $\mathrm{P}(A' \cap B' \cap C')$ in terms of the unknown probability $x = \mathrm{P}(B \cap C)$.

From earlier parts, $\mathrm{P}(A \cap B) = 0.1 \times 0.2 = 0.02$ and $\mathrm{P}(C) = c = 0.62$.
Let $x = \mathrm{P}(B \cap C)$. Since $A$ and $C$ are mutually exclusive, $A$ and $C$ do not overlap, meaning $\mathrm{P}(A \cap C) = 0$ and $\mathrm{P}(A \cap B \cap C) = 0$.
By considering the separate, non-overlapping regions in a Venn diagram:
$\mathrm{P}(A \text{ only}) = \mathrm{P}(A) - \mathrm{P}(A \cap B) = 0.1 - 0.02 = 0.08$
$\mathrm{P}(B \text{ only}) = \mathrm{P}(B) - \mathrm{P}(A \cap B) - \mathrm{P}(B \cap C) = 0.2 - 0.02 - x = 0.18 - x$
$\mathrm{P}(C \text{ only}) = \mathrm{P}(C) - \mathrm{P}(A \cap C) - \mathrm{P}(B \cap C) = 0.62 - 0 - x = 0.62 - x$
The probability of the region entirely outside all three sets is:
$\mathrm{P}(A' \cap B' \cap C') = 1 - [\mathrm{P}(A \text{ only}) + \mathrm{P}(A \cap B) + \mathrm{P}(B \text{ only}) + \mathrm{P}(B \cap C) + \mathrm{P}(C \text{ only})]$
$\mathrm{P}(A' \cap B' \cap C') = 1 - [0.08 + 0.02 + (0.18 - x) + x + (0.62 - x)]$
$\mathrm{P}(A' \cap B' \cap C') = 1 - (0.90 - x) = 0.10 + x$

2. Determine the valid range of values for $x = \mathrm{P}(B \cap C)$.

For the probabilities of the individual regions to be valid, they must be non-negative:
$\mathrm{P}(B \cap C) = x \ge 0$
$\mathrm{P}(B \text{ only}) = 0.18 - x \ge 0 \implies x \le 0.18$
$\mathrm{P}(C \text{ only}) = 0.62 - x \ge 0 \implies x \le 0.62$
Combining these conditions, the most restrictive bounds give the valid range for $x$ as:
$0 \le x \le 0.18$

3. Calculate the minimum and maximum possible values of $\mathrm{P}(A' \cap B' \cap C')$.

Substitute the minimum and maximum valid values of $x$ into the expression for $\mathrm{P}(A' \cap B' \cap C')$:
$\text{Minimum } \mathrm{P}(A' \cap B' \cap C') = 0.10 + 0 = 0.1$
$\text{Maximum } \mathrm{P}(A' \cap B' \cap C') = 0.10 + 0.18 = 0.28$