I know this is a very basic question, but my mathematics is coming out a bit wonky. Assume Events A, B are independent.
Define:
$Pr(A) = 1/6$
$Pr(B) = 1/4$
Let Event C = "at most one event out of A, B will occur".
So either A, B, or neither A nor B are all the possible outcomes for Event C.
$Pr(\neg A) = 5/6$
$Pr(\neg B) = 3/4$
$\neg A \cap \neg B = \neg A * \neg B = 5/8$
So, Event $C = Pr(A) + Pr(B) + Pr(\neg A \cap \neg B) = 1/6 + 1/4 + 5/8 = 25/24$
This probability is greater than 100%! I'm sure I've messed up somewhere. If it were 100%, that would exclude the possibility that both A and B could ever occur, which is definitely not the case.
UPDATE: exclude $Pr(A \cap B)$
Event $C = Pr(A) + Pr(B) – Pr( A \cap B) = 1/6 + 1/4 + 1/24 = 3/8$
Best Answer
Hint. The easy way to do it: the probability that at most one event will occur is the same as the probability that not both will occur, that is, $$1-P(A\cap B)\ .$$ Using the given information, you should easily be able to work this out.