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 $C = Pr(A) + Pr(B) - Pr( A \cap B) = 1/6 + 1/4 + 1/24 = 3/8$
$\endgroup$4 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$The correct formula is $$\Pr[A \cup B] = \Pr[A] + \Pr[B] - \Pr[A \cap B].$$ You have a flipped sign, which is why the probability you calculated is incorrect.
$\endgroup$ $\begingroup$AUB means either A event happens or event B event. A intersection B means both event A and B happend simultaneously.
$\endgroup$ 1 $\begingroup$$C$ is the event that exactly one of the following events:
- $A$ but not $B$
- $B$ but not $A$
- Not B and Not A (Neither A nor B)
1 is the event $A \setminus B$ or $A \cap B^C$
2 is the event $B \setminus A$ or $B \cap A^C$
3 is the event $B^C \cap A^C$ (not B and not A) or $(A \cup B)^C$ (not (A or B))
Observe that the events are mutually exclusive.
Hence,
$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$
Solution 1:
Recall that if $A$ and $B$ are independent, then so are the following pairs:
$$A, B^C$$
$$A^C, B$$
$$A^C, B^C$$
$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$
$$P(C) = P(A)P(B^C) + P(B)P(A^C) + P(B^C)P(A^C)$$
$$P(C) = P(A)(1-P(B)) + P(B)(1-P(A)) + (1-P(B))(1-P(A))$$
Solution 2:
$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$
$$P(C) = 1 - P(A \cap B)$$
$$P(C) = 1 - P(A)P(B)$$
$\endgroup$
