What happens to $P(A \cup B \cup C \cup D)$ when $B$ and $D$ are mutually exclusive? I know that 
$$P(A \cup B \cup C \cup D) = P(A)+P(B)+P(C)+P(D)\\-P(AB)-P(BC)-P(BD)-P(AD)-P(AC)-P(CD)\\+P(ABC)+P(ABD)+P(BCD)+P(CDA)-P(ABCD)$$ when $A$, $B$, $C$, and $D$ are not mutually exclusive. 
When does $P(A \cup B \cup C \cup D)$ achieve maximum possible value? Is it when $A$, $B$, $C$, and $D$ are mutually exclusive?
If $B$ and $D$ are mutually exclusive, then $P(BD), P(ABD), P(BCD), P(ABCD)$ are equal to $0$.  So 
$$P(A \cup B \cup C \cup D) = P(A)+P(B)+P(C)+P(D)\\
-P(AB)-P(BC)-P(AD)-P(AC)-P(CD)\\
+P(ABC)+P(CDA)
$$ 
Is this correct?
 A: "(Pairwise) mutually exclusive" would mean
$$
A\cap B = A\cap C = A \cap D = B\cap C = B\cap D = C \cap D = \varnothing. \tag 1
$$
Sometimes someone might use the term "mutually exclusive" to mean $A\cap B\cap C\cap D = \varnothing.$ That would happen if $A=\{1,2\},\  B=\{2,3\},\ C=\{3,4\},\  D= \{4,1\},$ and these are not pairwise mutually exclusive.
What is needed in order to maximize $P(A\cup B\cup C\cup D)$ subject to $P(A), P(B), P(C), P(D)$ having specified values is a somewhat weaker condition than $(1)$:
$$
P(A\cap B) = P(A\cap C) = P(A\cap D) = P(B\cap C) = P(B\cap D) = P(C\cap D) = 0. \tag 2
$$
For example, if $A = \{1,2,3\}$ and $B= \{3,4,5\}$ and $P(\{3\}) = 0$ then we would have $P(A\cap B)=0$ although $A\cap B\ne\varnothing.$
If $P(A\cap B)>0,$ then $P(A\cup B) < P(A)+P(B)$ and so
\begin{align}
& P(A\cup B\cup C\cup D) = P(A\cup B) + P(C\cup D) - P(A\cap B\cap C\cap D) \\[10pt]
< {} & P(A) + P(B) + P(C\cup D) \le P(A) + P(B) + P(C) + P(D),  
\end{align}
so $P(A\cup B\cup C\cup D)$ can be as big as $P(A) + P(B) + P(C) + P(D)$ only if $(2)$ is true.
Your last statement is correct.
