# set proof

• Aug 24th 2008, 03:22 PM
kid funky fried
set proof
Could someone please just give me a hint as to how I could prove this:

{(A∩B)’} = 1+P{AB} – P{A}– P{B}

Thanks.
• Aug 24th 2008, 03:48 PM
Plato
$\displaystyle \begin{array}{l} P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\ = 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\ = 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\ \end{array}$
• Aug 24th 2008, 06:48 PM
kid funky fried
Yes..
Yes, a big THANK YOU, Plato!

Kid
• Sep 2nd 2008, 01:36 AM
ban26ana
Quote:

Originally Posted by Plato
$\displaystyle \begin{array}{l} P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\ = 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\ = 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\ \end{array}$

Can you (or anyone) explain what rules or laws got you to this? I follow the first step, because it is DeMorgan's Law. But from there I'm lost. I'm looking through all my notes and my texbook, but I'm not seeing the rules that allow you to make those changes.
• Sep 2nd 2008, 01:54 AM
Moo
Quote:

Originally Posted by ban26ana
Can you (or anyone) explain what rules or laws got you to this? I follow the first step, because it is DeMorgan's Law. But from there I'm lost. I'm looking through all my notes and my texbook, but I'm not seeing the rules that allow you to make those changes.

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 $\displaystyle A \cap B$

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 AM
ban26ana
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 AM
Moo
Quote:

Originally Posted by ban26ana
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)?

Yes, because $\displaystyle A' \cap B'=(A \cup B)'$, which is de Morgan's rule.
• Sep 2nd 2008, 02:17 AM
ban26ana
Thanks again Moo. I get it!