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Interesting Probability question. I have too proof that

\begin{align*} P(A \cup B \cup C) &= P(A) + P(B) + P(C) \\ &\quad - P(A \cap B) - P(B \cap C) - P(A \cap C) \\ &\quad + P(A \cap B \cap C). \end{align*}

Using probability introductory formulas and theorems. I understand fully when using the Venn diagram but I'm having trouble proving it mathematically.

I have tried the following but haven't been able to continue.

\begin{align*} P(A \cup B \cup C) &= P(A \cup (B \cup C) \tag{1} \\ &= P(A)+ P((B \cup C) \cap A')\tag{2} \\ &= P(A)+ P((B \cap A') \cup (C \cap A'))\tag{3} \\ &= P(A)+ P(B \cap A')+ P(C \cap A')\tag{4} \end{align*}

We also know that

\begin{align*} \begin{split} P(B)&=P(A \cap B)+P(B \cap A'), \\ P(C)&=P(A \cap C)+P(C \cap A'). \end{split} \tag{5} \end{align*}

Thus

\begin{align*} P(A \cup B \cup C) = P(A)+ P(B)- P(A \cap B)+ P(C)-P(C \cap B) \tag{6} \end{align*}

So I'm almost there but to complete the equation I'm still missing $-P(A \cap C)$ and $P(A \cap B \cap C)$.

I was thinking that $P(A \cup B \cup C)= 1- P(A' \cup B' \cup C')$ but I'm not sure how this is helpful.

Any help is appreciated.

Edit: I was also successful in proving $P(A \cup B)= P(A)+P(B)-P(A \cup B)$, and I'm trying to use the same techniques to proof this statement.

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From the third step to the fourth step, you did not include any intersection term.

\begin{align} &P(A) + P( B \cap A') \cup (C \cap A') \\&=P(A)+ P(B \cap A') + P(C \cap A') \color{blue}{- P(B \cap C \cap A')}\\ &=P(A)+P(B)-P(B \cap A) + P(C)-P(C \cap A) -(P(B\cap C)-P(A \cap B \cap C)) \end{align}

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  • $\begingroup$ Just to clarify... how are you going from the blue segment to −(P(B∩C)−P(A∩B∩C))?? Thanks for your help btw $\endgroup$ – boby Sep 24 '18 at 1:45
  • $\begingroup$ $P((B \cap C) \cap A') + P((B \cap C) \cap A) = P(B \cap C)$ $\endgroup$ – Siong Thye Goh Sep 24 '18 at 1:46
  • $\begingroup$ Shouldn't it be P((B∩C)∩A′) - P((B∩C)∩A)=P(B∩C)?? $\endgroup$ – boby Sep 24 '18 at 1:49
  • $\begingroup$ If you let $D = B \cap C$, we get $P(D \cap A') +P(D \cap A) = P(D)$. That should be a plus. $\endgroup$ – Siong Thye Goh Sep 24 '18 at 1:50
  • $\begingroup$ ohhh is see... how to you manage to get (positive) P(AnBnC) as seen in the statement though? because you need to add that intersection after subtracting all the others. $\endgroup$ – boby Sep 24 '18 at 1:53
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Use the "change of variables" $B\cup C=D$ and everything follows.

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