# How to remember these probability results?

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)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-2P(A \cap B \cap C)$

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

## 1 Answer

A picture is worth a thousand words

• What you have labelled as $A\cap B$ should really be $A\cap B\cap \bar{C}$ and similarly for the two other regions – David Quinn Aug 28 '15 at 17:26
• 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 – Giovanni Aug 28 '15 at 17:33
• Yes I mean the complement of C. $A\cap B$ should include $A\cap B\cap C$ – David Quinn Aug 28 '15 at 17:36
• As I said, it does contain the triple intersection. – Giovanni Aug 28 '15 at 17:41