1. Tardigrade
  2. Question
  3. Mathematics
  4. Let 0< P(A)< 1,0< P(B)< 1 P(A ∪ B)=P(A)+P(B)-P(A) . P(B), then:

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Q. Let $0< P(A)< 1,0< P(B)< 1 \& P(A \cup B)=P(A)+P(B)-P(A) . P(B)$, then:

Probability - Part 2

Solution:

$P \left(( A \cup B )^{ c }\right)= P \left( A ^{ c }\right) \cdot P \left( B ^{ c }\right)$