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We subtract the intersection because adding the individual probabilities $P(E)$ and $P(F)$ counts the overlapping region twice. Subtracting $P(E \cap F)$ once corrects this double-counting so that every outcome in the union is counted exactly once.

Conditions: E and F are overlapping events in a probability space.

We subtract the intersection because adding the individual probabilities $P(E)$ and $P(F)$ counts the overlapping region twice. Subtracting $P(E \cap F)$ once corrects this double-counting so that every outcome in the union is counted exactly once.

Conditions: E and F are overlapping events in a probability space.