Given two events, A and B, what is the probability that at most one will occur? 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 $C = Pr(A) + Pr(B) - Pr( A \cap  B) = 1/6 + 1/4 + 1/24 = 3/8$
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.
$C$ 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:

$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$

$$P(C) = 1 - P(A \cap B)$$

$$P(C) = 1 - P(A)P(B)$$