The two-event addition rule generalises, but the signs alternate: add the singles, subtract every pair, then add the triple back because it was removed one time too many.

$$P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A\cap B)-P(A\cap C)-P(B\cap C) + P(A\cap B\cap C)$$

Return $P(A \cup B \cup C)$.

solve(0.5, 0.5, 1/3, 0.25, 1/6, 1/6, 1/6)  ->  0.9166666666666666

Sketch a three-circle diagram if the signs feel arbitrary: the centre region sits inside all three singles and all three pairs.