If $A,B$ and $C$ are $3$ events, then

  1. $P$(Exactly one of $A,B,C$ occurs)$=P(A)+P(B)+P(C)-2[P(A \cap B)+P(B \cap C)+P(A \cap C)]+3P(A \cap B \cap C)$

  2. $P$(Exactly two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-3P(A \cap B \cap C)$

  3. $P$(At least two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-2P(A \cap B \cap C)$

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    $\begingroup$ Here is how I remember them: math.stackexchange.com/questions/151635/… $\endgroup$ – user940 Aug 28 '15 at 16:39
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    $\begingroup$ yeah, should be $-2P(A\cap B\cap C)$ $\endgroup$ – Giovanni Aug 28 '15 at 16:50
  • $\begingroup$ @Henry Thankx Done editing. $\endgroup$ – Jack Aug 28 '15 at 16:53
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    $\begingroup$ @Giovanni Thankx $\endgroup$ – Jack Aug 28 '15 at 16:53

A picture is worth a thousand words

  • $\begingroup$ What you have labelled as $A\cap B$ should really be $A\cap B\cap \bar{C}$ and similarly for the two other regions $\endgroup$ – David Quinn Aug 28 '15 at 17:26
  • $\begingroup$ Hi David, I am sorry but I really don't understand your comment. I don't think that the closure of a set has any meaning in this context, so I guess you meant the complement of $C$. If that is the case I am pretty sure that whoever did this (I took it from wikipedia) wanted the whole intersection of $A$ and $B$ there.Of course it includes the part where they all intersect, but this is fairly obvious :) Please, let me know if I misunderstood your comment $\endgroup$ – Giovanni Aug 28 '15 at 17:33
  • $\begingroup$ Yes I mean the complement of C. $A\cap B$ should include $A\cap B\cap C$ $\endgroup$ – David Quinn Aug 28 '15 at 17:36
  • $\begingroup$ As I said, it does contain the triple intersection. $\endgroup$ – Giovanni Aug 28 '15 at 17:41

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