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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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The event is $(A \cup B)^c$. The probability is $1 - P(A \cup B)$. Events are sets, probabilities are numbers. – Nate Eldredge Sep 7 '12 at 1:51
I was under the impression that Pr(A∪B)c = 1 - Pr(A∪B).. – James Sep 7 '12 at 1:55
That isn't the usual notation, but perhaps your book uses it. – Nate Eldredge Sep 7 '12 at 1:57
Maybe that is it. My professor used this notation in lecture. It might have been informal. – James Sep 7 '12 at 1:58
"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 laws – Dilip Sarwate Sep 7 '12 at 4:09
up vote 1 down vote accepted

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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would Pr(A∩B)c equate to (1-Pr(A)) + (1-Pr(B)) - (1-Pr(A∩B))? – James Sep 7 '12 at 2:17
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 $(A^c \cup B^c)^c = A\cap B$. – Dilip Sarwate Sep 7 '12 at 4:06

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