已知事件 A 和 B 满足 $P(A \cup B') = \frac{2}{3}$ 且 $P(A) = \frac{1}{4}$。

1. Express the required conditional probability using the standard formula and simplify the numerator.

$P(A \mid A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$\nSince event $A$ is a subset of $A \cup B$, their intersection is simply $A$:\n$P(A \mid A \cup B) = \frac{P(A)}{P(A \cup B)}$

2. Determine $P(A \cup B)$ using the given probabilities and set complements.

Using De Morgan's laws, the complement of $A \cup B'$ is $A' \cap B$. Therefore:\n$P(A' \cap B) = 1 - P(A \cup B') = 1 - \frac{2}{3} = \frac{1}{3}$\nSince $A$ and $A' \cap B$ are mutually exclusive and their union forms $A \cup B$:\n$P(A \cup B) = P(A) + P(A' \cap B) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}$

3. Evaluate the final conditional probability.

$P(A \mid A \cup B) = \frac{1/4}{7/12} = \frac{1}{4} \times \frac{12}{7} = \frac{3}{7}$

1. Apply the addition rule and the definition of independent events to express $P(A \cup B')$ in terms of $P(A)$ and $P(B')$.

By the addition rule of probability:
$P(A \cup B') = P(A) + P(B') - P(A \cap B')$
Since $A$ and $B$ are independent, $A$ and $B'$ are also independent. Therefore, $P(A \cap B') = P(A)P(B')$.
$P(A \cup B') = P(A) + P(B') - P(A)P(B')$

2. Substitute the given values and solve for $P(B')$.

$\frac{2}{3} = \frac{1}{4} + P(B') - \frac{1}{4}P(B')$
$\frac{2}{3} - \frac{1}{4} = \frac{3}{4}P(B')$
$\frac{5}{12} = \frac{3}{4}P(B')$
$P(B') = \frac{5}{12} \times \frac{4}{3} = \frac{5}{9}$

3. Find $P(B)$ using the complement rule.

$P(B) = 1 - P(B')$
$P(B) = 1 - \frac{5}{9} = \frac{4}{9}$

1. Express $P(A \cup B')$ in terms of $P(B)$ and $P(A \cap B)$.

Using the addition rule for probabilities:
$P(A \cup B') = P(A) + P(B') - P(A \cap B')$
Since the part of $A$ not in $B$ is $P(A \cap B') = P(A) - P(A \cap B)$, this simplifies to:
$P(A \cup B') = P(A) + P(B') - (P(A) - P(A \cap B))$
$P(A \cup B') = P(B') + P(A \cap B)$
Substituting the given values and $P(B') = 1 - P(B)$:
$\frac{2}{3} = 1 - P(B) + P(A \cap B)$
$P(B) = \frac{1}{3} + P(A \cap B)$

2. Determine the basic range of $P(B)$ using the minimum and maximum possible values of $P(A \cap B)$.

The probability of an intersection must satisfy $0 \leqslant P(A \cap B) \leqslant P(A)$.
Since $P(A) = \frac{1}{4}$, we have:
$0 \leqslant P(A \cap B) \leqslant \frac{1}{4}$
Substitute $P(A \cap B) = P(B) - \frac{1}{3}$ into the inequality:
$0 \leqslant P(B) - \frac{1}{3} \leqslant \frac{1}{4}$
$\frac{1}{3} \leqslant P(B) \leqslant \frac{1}{3} + \frac{1}{4}$
$\frac{1}{3} \leqslant P(B) \leqslant \frac{7}{12}$

3. Exclude the value of $P(B)$ that would make $A$ and $B$ independent.

If events $A$ and $B$ were independent, we would have $P(A \cap B) = P(A)P(B) = \frac{1}{4}P(B)$.
Substituting this into the equation from Step 1 (or retrieving the answer calculated in part ii):
$P(B) = \frac{1}{3} + \frac{1}{4}P(B)$
$\frac{3}{4}P(B) = \frac{1}{3} \implies P(B) = \frac{4}{9}$
Since we are given that $A$ and $B$ are not independent, $P(B) \neq \frac{4}{9}$.
Combining this with the continuous bounds from Step 2, the range of possible values is:
$\frac{1}{3} \leqslant P(B) < \frac{4}{9} \quad \text{or} \quad \frac{4}{9} < P(B) \leqslant \frac{7}{12}$