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    So far I've done the following:

    Let n=2 then we get \displaystyle \mathbb{P} \left( \bigcup_{i=1}^2 A_i \right) =\mathbb{P} \left( A_1 \cup A_2 \right) = \mathbb{P}(A_1)+\mathbb{P}(A_2)-\mathbb{P}(A_1 \cap A_2) on the LHS while I get \displaystyle \sum_{i=1}^2 \mathbb{P}(A_i) - \mathbb{P}(A_1 \cap A_2) on the RHS which equals the LHS.

    Then say it's true for n=a hence I get

    \displaystyle \mathbb{P} \left( \bigcup_{i=1}^a A_i \right) = \sum_{i=1}^{a}\mathbb{P}(A_i) - \sum_{i<j} \mathbb{P}(A_i \cap A_j) + \sum_{i<j<k} \mathbb{P}(A_i \cap A_j \cap A_k)-...+(-1)^{a+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a)

    So then let n=a+1 and we get:

    \displaystyle \mathbb{P} \left( \bigcup_{i=1}^{a+1} A_i \right)= \mathbb{P} \left( \bigcup_{i=1}^a A_i \cup A_{a+1} \right) = \mathbb{P} \left( \bigcup_{i=1}^a A_i \right) + \mathbb{P}(A_{a+1})-\mathbb{P} \left( \bigcup_{i=1}^a A_i  \cap A_{a+1}\right)

    which turns into

    \displaystyle \sum_{i=1}^{a+1}\mathbb{P}(A_i) - \sum_{i<j} \mathbb{P}(A_i \cap A_j) + \sum_{i<j<k} \mathbb{P}(A_i \cap A_j \cap A_k)-...+(-1)^{a+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a) - \mathbb{P} \left( \bigcup_{i=1}^a A_i  \cap A_{a+1}\right)

    and now I don't understand how I can turn the last 2 terms into (-1)^{(a+1)+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a \cap A_{a+1})
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    (Original post by RDKGames)


    So far I've done the following:

    [...]

    which turns into

    \displaystyle \sum_{i=1}^{a+1}\mathbb{P}(A_i) - \sum_{i<j} \mathbb{P}(A_i \cap A_j) + \sum_{i<j<k} \mathbb{P}(A_i \cap A_j \cap A_k)-...+(-1)^{a+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a) - \mathbb{P} \left( \bigcup_{i=1}^a A_i  \cap A_{a+1}\right)

    and now I don't understand how I can turn the last 2 terms into (-1)^{(a+1)+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a \cap A_{a+1})
    What probability space axioms are you working with? One of the most fundamental is that if A and B are disjoint events (empty intersection) then  \mathbb{P}(A \cup B)=\mathbb{P}(A)+\mathbb{P}(B) . But in your proof, the A_i need not be disjoint. So it looks like you need to do some complementing. Note that
     \mathbb{P}((\cup A_i)^c) = \mathbb{P}(\cap A^c_i)

    EDIT: In your argument above, the indices i<j and i<j<k run only upto a, NOT upto a+1, so you actually need to show something more complicated than that the last two terms turn into (-1)^{(a+1)+1} \mathbb{P}(A_1 \cap A_2 \cap ... \cap A_a \cap A_{a+1}) (which I imagine is probably false in general).

    A good idea is to note that \mathbb{P} \left( \bigcup_{i=1}^a A_i \cap A_{a+1}\right)=\mathbb{P}(\cup (A_i \cap A_{a+1}) is a union of a events so you can use the inductive hypothesis again.
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    (Original post by RDKGames)
    which turns into
    I'd have to Google it myself, but I am wary of your working.

    When you say "which turns into", what are the limits on the summation? E.g i<j. Is it i&lt;j\leq a, or i&lt;j\leq a+1. I suspect the former, though I could be misreading it.

    Edit: I see theOldBean has now picked that up.
 
 
 
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