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Using the inclusion exclusion principle - http://www.proofwiki.org/wiki/Inclusion-Exclusion_Principle - if I set $n=2$ I get the following -

$$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) + P(A_1 \cap A_2)$$

when the correct answer should be -

$$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)$$

I have the term $P(A_1 \cap A_2)$ at the end the first equation due to the last part of the inclusion exclusion principle - $$(-1)^{n-1}P(\cap_i^n A_i)$$

It seems that I shouldn't be including that if I want to have the correct answer...but surely I have to include it as I can't just drop an arbitrary term from some formula...so what am I missing?

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  • $\begingroup$ The set $A$ is not in the union on the left, so should not appear on the right side. $\endgroup$
    – coffeemath
    Nov 30, 2012 at 12:41

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In both of the places where you wrote $A$ you probably mean $A_1$; I’m going to assume so.

When $n=2$ the last term is

$$(-1)^{n-1}P\left(\bigcap_{i=1}^nA_i\right)=(-1)^1P\left(\bigcap_{i=1}^2A_i\right)=-P(A_1\cap A_2)\;,$$

just as it should be. Somehow you added an extra term that is not present in the expression on the cited Proof Wiki page.

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  • $\begingroup$ Oops, well then I would have $$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) - P(A_1 \cap A_2)$$. If the last term is correct then it is the second last term in my expression that is incorrect. I am getting that term $- P(A_1 \cap A_2)$ from the second term of the inclusion exclusion principle $\sum_{i<j}^n P(A_i \cap A_j)$. Have I made a mistake by including this term? $\endgroup$
    – csss
    Nov 30, 2012 at 12:48
  • $\begingroup$ @csss: No, you still have one term too many: you’ve written the last term twice. $\endgroup$
    – Brian M. Scott
    Nov 30, 2012 at 13:03
  • $\begingroup$ I have wrote it twice because when I let $n=2$ that is what the inclusion exclusion formula gives me. I get $-P(A_1 \cap A_2)$ twice. Once from the 2nd term of the IE formula and once from the last term of the IE formula. How come you don't get two $-P(A_1 \cap A_2)$ terms? $\endgroup$
    – csss
    Nov 30, 2012 at 13:15
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    $\begingroup$ @csss: You’re not reading the formula correctly. When $n=2$ the only terms are $\sum_{i=1}^2P(A_i)$ and $-\sum_{1\le i<j\le 2}P(A_i\cap A_j)$. That second term is the term $(-1)^{2-1}P\left(\bigcap_{i=1}^2A_i\right)$. $\endgroup$
    – Brian M. Scott
    Nov 30, 2012 at 13:24

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