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We are given events A, B. If "the probability of at least one of A and B happens" translates to Pr($A$$B$), and "at most one of A and B happens" is the complement of the former, would the probability be Pr($A$$B$)c?

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    "at most one of $A$ and $B$ happens" is the same as "at least one of $A^c$ and $B^c$ happens" and so the event is $A^c \cup B^c$ which can also be expressed as $(A\cap B)^c $ via DeMorgan's laws2012-09-07

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The probability that at most one of $A$ or $B$ happens can be broken down into 3 disjoint cases: neither $A$ nor $B$ happens, $A$ but not $B$ happens, and $B$ but not $A$ happens. This is: $Pr[(A \cup B)^c \cup (A \cap B^c) \cup (A^c \cap B)]$ $=Pr[(A \cap B^c) \cup (A^c \cap B^c) \cup (A^c \cap B)]$ $=Pr[(A \cup A^c) \cap B^c \cup A^c \cap(B^c \cup B)]$ $=Pr[B^c \cup A^c]$ $=Pr[(A\cap B)^c]$

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    More simply, the probability that at most one of $A$ and $B$ happens is the probability that _at least one_ of $A^c$ and $B^c$ happens, and so the event is $A^c \cup B^c$ and has probability $P(A^c \cup B^c) = 1 - P(A\cap B)$ since DeMorgan's laws tell us that $(A^c \cup B^c)^c = A\cap B$.2012-09-07