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Given $P(x \lor y) = P(x) + P(y)$ ; if $x$ and $y$ are disjoint

How to prove the following: $$P(x \lor y) = P(x) + P(y) - P(x \land y) \qquad\text{if }x\text{ and }y\text{ are not disjoint.}$$

Is the following also correct? $$P(x)= P(x') + P(x \land y)$$ where $x = x' \lor (x \land y$)

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Draw a Venn diagram and cross-hatch $A$ as ///, $B$ as \\\ and notice that $A\cap B$ is cross-hatched both ways and so gets counted twice in $P(A)+P(B)$. So subtract $P(A\cap B)$ from $P(A) + P(B)$ to get $P(A\cup B)$. –  Dilip Sarwate Mar 9 '12 at 13:59

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up vote 1 down vote accepted

Instead of using "and", I'll use the intersection symbol "$\cap$"; likewise I'll use the union symbol "$\cup$" for "or".

Your last formula isn't correct. $x'$ is everything not in $x$ (the complement of $x$). Instead, you want to use $x\setminus y$, which is defined to be everything that is in $x$ but not in $y$.

Your last formula should be $x=(x\setminus y )\cup (x\cap y)$. This will be a disjoint union; so: $$ P(x)=P(x\setminus y)+P(x\cap y), $$ or $$\tag{1} P(x\setminus y)= P(x)-P(x\cap y). $$

Now, for your original problem, the strategy for finding the formula for $P(x\cup y)$ is write the event $x\cup y$ as the union of disjoint events and use the formula you have for the probability of disjoint events (note the formula generalizes to three mutually disjoint events).

One way to do this is to write $$\tag{2} x\cup y= (x\setminus y)\cup (y\setminus x)\cup (x\cap y) $$ (note the three events on the right are mutually disjoint; drawing a Venn diagram for the two sets $x$ and $y$ should help to see why $(2)$ holds). Using equations $(1)$, $(2)$, and the probability formula for the union of disjoint events, you should be able to now prove that $$ P(x \cup y)=P(x)+P(y)-P(x\cap y). $$

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Thanks for the explanation. I can understand it now. –  m-lala Mar 9 '12 at 14:38

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