Answer
$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(B\cap C)-P(A\cap B)-P(A\cap C)+P(A\cap B\cap C)$
Work Step by Step
$P(A\cup B\cup C)=P[A\cup (B\cup C)]=P(A)+P(B\cup C)-P[A\cap(B\cup C)]$
But:
$P(B\cup C)=P(B)+P(C)-P(B\cap C)~~$ which leads us to:
$P[A\cap(B\cup C)]=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)$
Returning to the initial formula:
$P(A\cup B\cup C)=P[A\cup (B\cup C)]=P(A)+P(B\cup C)-P[A\cap(B\cup C)]=P(A)+P(B)+P(C)-P(B\cap C)-[P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)]=P(A)+P(B)+P(C)-P(B\cap C)-P(A\cap B)-P(A\cap C)+P(A\cap B\cap C)$
