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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 $C = Pr(A) + Pr(B) - Pr( A \cap B) = 1/6 + 1/4 + 1/24 = 3/8$

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4 Answers 4

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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.

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  • $\begingroup$ Great, so if we were modeling the event C with a tree, normally we use the Sum rule to add the probabilities of all the outcomes that satisfy the event. How would we accommodate a subtraction, as I think this would imply? Is there a way to make the third branch $\neg Pr(A \cap B)$ into something that we would add to get the Pr(C)? $\endgroup$
    – compguy24
    Commented Apr 22, 2014 at 0:39
  • $\begingroup$ You could use $P(C)=P(A\cap\neg B)+P(B\cap\neg A)+P(\neg A\cap\neg B)$. $\endgroup$
    – David
    Commented Apr 22, 2014 at 0:51
  • $\begingroup$ So, if my update is correct (see above), I should expect $Pr(C) = 3/8$. If I use your suggested formula, I get $(1/6 * 3/4) + (1/4 * 5/6) + (5/6 * 3/4) = 23/24 \neq 3/8)$. Have I not followed your suggestion correctly? $\endgroup$
    – compguy24
    Commented Apr 22, 2014 at 1:12
  • $\begingroup$ Sorry but I do not understand at all why you are calculating $P(A)+P(B)-P(A\cap B)$. This is the probability that at least one of the events occurs and I can't see what that has to do with your problem. I guess I must have misread the question. $\endgroup$
    – David
    Commented Apr 22, 2014 at 1:26
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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.

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AUB means either A event happens or event B event. A intersection B means both event A and B happend simultaneously.

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  • $\begingroup$ Please, use MathJax (i.e. LaTeX commands) for mathematical formulas. $\endgroup$
    – Taroccoesbrocco
    Commented Sep 13, 2017 at 16:35
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$C$ is the event that exactly one of the following events:

  1. $A$ but not $B$
  2. $B$ but not $A$
  3. 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:

  1. $$A, B^C$$

  2. $$A^C, B$$

  3. $$A^C, B^C$$

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

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

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

Solution 2:

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

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

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

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