1. ## 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.

2. $\begin{array}{l}
P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\
= 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\
= 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\
\end{array}$

3. ## Yes..

Yes, a big THANK YOU, Plato!

Kid

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

5. 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 :

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

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

This is why $P(A^c \cup B^c)=P(A^c)+P(B^c)-P(A^c \cap B^c)$
$^c$ denotes the complement - see here : Complement (set theory - Wikipedia, the free encyclopedia)
And we have $P(A^c)=1-P(A)$

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

...

6. 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)?

7. 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 $A' \cap B'=(A \cup B)'$, which is de Morgan's rule.

8. Thanks again Moo. I get it!