Two events $A$ and $B$ are such that $P(A)=p$ and $P(B)=\frac{5}{4}p$. It is given that $P(A\cup B)=P(A\mid B)=0.6$.

1. Express $P(A \cap B)$ in terms of $p$ using the conditional probability formula.

$P(A \cap B) = P(A \mid B)P(B)$
$P(A \cap B) = 0.6 \left(\frac{5}{4}p\right) = 0.75p$

2. Form an equation in terms of $p$ using the addition rule.

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.6 = p + 1.25p - 0.75p$

3. Solve the linear equation to find $p$.

$0.6 = 1.5p$
$p = \frac{0.6}{1.5}$
$p = 0.4$

1. Use the value of $p$ from part (a) to find $P(A)$, $P(B)$, and $P(A \cap B)$.

From part (a), $p = 0.4$.
$P(A) = 0.4$
$P(B) = \frac{5}{4}(0.4) = 0.5$
$P(A \cap B) = P(A \mid B)P(B) = 0.6 \times 0.5 = 0.3$

2. Calculate $P(A' \cap B)$ and $P(A')P(B)$ to test for independence.

$P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.3 = 0.2$
$P(A')P(B) = (1 - P(A))P(B) = (1 - 0.4)(0.5) = 0.6 \times 0.5 = 0.3$

Since $P(A' \cap B) \neq P(A')P(B)$, the events $A'$ and $B$ are not independent.

1. Establish the maximum possible value for $q$ using the probability of $A$ found in part (a).

From part (a), we calculated $P(A) = 0.4$.
The intersection of two events is a subset of each individual event, so its probability cannot exceed the probability of either event. Therefore, $P(A \cap C) \le P(A)$.
Given $P(A \cap C) = q$, we have:
$q \le 0.4$

2. Express the probability of the union of $C$ and $D$ in terms of $q$.

Using the addition rule of probability for events $C$ and $D$:
$P(C \cup D) = P(C) + P(D) - P(C \cap D)$
Substitute the given values $P(C) = 0.55$, $P(D) = 0.6$, and $P(C \cap D) = q$:
$P(C \cup D) = 0.55 + 0.6 - q = 1.15 - q$

3. Relate $P(A \cup C \cup D)$ to $P(C \cup D)$ to find the minimum value.

Since the union of $C$ and $D$ is a subset of the union of $A$, $C$, and $D$ (i.e., $C \cup D \subseteq A \cup C \cup D$), its probability provides a lower bound:
$P(A \cup C \cup D) \ge P(C \cup D)$
$P(A \cup C \cup D) \ge 1.15 - q$
To find the minimum possible value for $P(A \cup C \cup D)$, we must subtract the largest possible value of $q$. Substituting the maximum value $q = 0.4$:
$\text{Minimum } P(A \cup C \cup D) = 1.15 - 0.4 = 0.75$
This minimum is achieved when $q = 0.4$.