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The union of two sets $A$ and $B$ is the set obtained by combining the members of each. This is written $A\cup B$, and is pronounced ``$A$ union $B$'' or ``$A$ cup $B$.'' The union of sets $A_1$ through $A_n$ is written $\bigcup_{i=1}^n A_i$.

Let $A$, $B$, $C$, ... be sets, and let $P(S)$ denote the probability of $S$. Then

P(A\cup B) = P(A)+P(B)-P(A\cap B).
\end{displaymath} (1)

$P(A\cup B\cup C) = P[A\cup(B\cup C)]$
$= P(A)+P(B\cup C)-P[A\cap(B\cup C)]$
$= P(A)+[P(B)+P(C)-P(B\cap C)]$
$\quad -P[(A\cap B)\cup(A\cap C)]$
$= P(A)+P(B)+P(C)-P(B\cap C)$
$\quad -\{P(A\cap B)+P(A\cap C)-P[(A\cap B)\cap (A\cap C)]\}$
$= P(A)+P(B)+P(C)-P(A\cap B)$
$\quad -P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).$ (2)
If $A$ and $B$ are Disjoint, by definition $P(A\cap B) = 0$, so
P(A\cup B) = P(A)+P(B).
\end{displaymath} (3)

Continuing, for a set of $n$ disjoint elements $E_1$, $E_2$, ..., $E_n$
P\left({\,\bigcup_{i=1}^n E_i}\right)= \sum_{i=1}^n P(E_i),
\end{displaymath} (4)

which is the Countable Additivity Probability Axiom. Now let
E_i \equiv A\cap B_i,
\end{displaymath} (5)

P\left({\,\bigcup_{i=1}^n E\cap B_i}\right)= \sum_{i=1}^n P(E\cap B_i).
\end{displaymath} (6)

See also Intersection, Or

© 1996-9 Eric W. Weisstein