Could someone please just give me a hint as to how I could prove this:

{(A∩B)’} = 1+P{A∪B} – P{A}– P{B}

Thanks.

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- Aug 24th 2008, 03:22 PMkid funky friedset proof
Could someone please just give me a hint as to how I could prove this:

{(A∩B)’} = 1+P{A∪B} – P{A}– P{B}

Thanks. - Aug 24th 2008, 03:48 PMPlato
$\displaystyle \begin{array}{l}

P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\

= 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\

= 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\

\end{array}$ - Aug 24th 2008, 06:48 PMkid funky friedYes..
Yes, a big THANK YOU, Plato!

Kid - Sep 2nd 2008, 01:36 AMban26ana
- Sep 2nd 2008, 01:54 AMMoo
There's a property (I don't know the name...) that says :

$\displaystyle P(A \cup B)=P(A)+P(B)-P(A \cap B)$

You can see it on a Venn diagram. If you add A and B, you'll add twice $\displaystyle A \cap B$

http://upload.wikimedia.org/wikipedi..._A_union_B.png

This is why $\displaystyle P(A^c \cup B^c)=P(A^c)+P(B^c)-P(A^c \cap B^c)$

$\displaystyle ^c$ denotes the complement - see here : Complement (set theory - Wikipedia, the free encyclopedia)

And we have $\displaystyle P(A^c)=1-P(A)$

So :

$\displaystyle P(A^c)+P(B^c)-P\underbrace{(A^c \cap B^c)}_{(A \cup B)^c ~, ~\text{by de Morgan's law}}=(1-P(A))+(1-P(B))-(1-P(A \cup B))$

... :p - Sep 2nd 2008, 02:04 AMban26ana
Thanks Moo. I found the theorem in my book that says P(A) = 1 - P(A'). Is that the same one that makes P(A' *intersect* B') = 1 - P(A U B)?

- Sep 2nd 2008, 02:15 AMMoo
- Sep 2nd 2008, 02:17 AMban26ana
Thanks again Moo. I get it!