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Addition of probabilities

Addition theorem for mutually exclusive events
If AA and BB are mutually exclusive (AB=A\cap B=\varnothing), then P(AB)=P(A)+P(B).P(A\cup B)=P(A)+P(B). Interpretation: favorable outcomes for ABA\cup B do not overlap, so their counts add — and under the classical scheme the fractions m/nm/n add as well. Special case: P(A)+P(A)=1P(A)+P(\overline{A})=1.
Without mutual exclusivity you cannot plug this formula in “as is”
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