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Truthfully, this is a homework problem. I've come across a solution, but I'm really trying to figure out how this works, hopefully at an intuitive level.

We have four events, A, B, C and D, and we are trying to find the probability that exactly one event occurs.

I've seen the "Inclusion exclusion principle", and that would help if I had to find the probability of A or B or C or D. I feel this should be trivial, probability is my weakest math, which is why I'm taking it, but this particular problem is giving me a lot of trouble. Any help would be greatly appreciated.

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Also posted on reddit.com/r/math. –  Alex Becker Sep 15 '12 at 0:19
    
please add the homework tag. Thanks. –  Seyhmus Güngören Sep 15 '12 at 0:30
    
Yes, I was posted on reddit.com/r/math. Its amazing what a small world the internet has become.. It seems the "homework" tag was added to the upper right hand corner, is that sufficient? This is my first time using *.stackexchange.com, I am not sure how to change the tag. I apologize. –  Anthony Sep 15 '12 at 0:53
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3 Answers

There are many possible formulas. The appropriate one depends on the information you are given. I assume that you know the basic Inclusion Exclusion Principle for computing $\Pr(A\cup B\cup C\cup D)$. A modification of that idea will take care of your problem.

When we find $$\Pr(A)+\Pr(B)+\Pr(C)+\Pr(D),$$ the situations in which two of $A$, $B$, $C$, and $D$ occur have been taken account of twice. But we want to count them zero times. So we want to subtract $$2\left(\Pr(A\cap B)+\Pr(A\cap C+\Pr(A\cap D)+\Pr(B\cap C)+\Pr(B\cap D)+\Pr(C\cap D)\right).$$ But we have been overenthusiastic in our subtractions, for we subtracted too often the probability that three of $A$, $B$, $C$, $D$ occur. Use a picture (Venn Diagram) to decide what we must add back. And there will be also an issue about the probability that all four of $A$, $B$, $C$, and $D$ occur.

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+1 good hints for a homwork problem without providing a complete answer. –  Michael Chernick Sep 15 '12 at 2:17
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Decompose the event into the 4 simpler events $A\cap B^c\cap C^c\cap D^c$, $A^c\cap B\cap C^c\cap D^c$, $A^c\cap B^c\cap C\cap D^c$, and $A^c\cap B^c\cap C^c\cap D$. These are disjoint, so you can simply sum the probabilities of each and you'll be done.

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Thank you for your response. I have actually managed to understand this up to that point. At least visually. I was under the (perhaps false) assumption that the answer would look something like the result of the inclusion-exclusion principle. Along the lines of P(A)+P(B)... perhaps - 2(AB) - 2(AC).... As I mentioned, probability is my achilles heel of math. Perhaps I am missing a fundamental on how to expand this? Or why it does not need to be expanded further? –  Anthony Sep 15 '12 at 0:44
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Using ${}^c$ to denote complement of an event, "exactly one of $A,B,C,D$" means $AB^cC^cD^c \cup A^c B C^c D^c \cup A^c B^c C D^c \cup A^c B^c C^c D$, those four being mutually exclusive. So one way to write it is

$$P(AB^cC^cD^c) + P(A^c B C^c D^c) + P(A^c B^c C D^c) + P(A^c B^c C^c D) $$

If you want to express it without any complements, note that e.g. $P(A B^c C^c D^c) = P(A) - P(AB \cup AC \cup AD)$, and use inclusion-exclusion on $P(AB \cup AC \cup AD)$ (and similarly for the other three).

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Ah! Thank you Mr. Israel ! Would you believe me if I told you I couldn't find this anywhere on the internet or in a book for at least 7 solid hours of search? I was able to minimize the problem to the complement form on my own. And I do believe your expression of the answer without complements is exactly what I am looking for. Its sad, that answer seems so obvious, and I struggled so much with it. Is there any resources, or information you would suggest to help solidify this principle? –  Anthony Sep 15 '12 at 0:50
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