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.
$\displaystyle \begin{array}{l}
P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\
= 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\
= 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\
\end{array}$
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 :
$\displaystyle P(A^c)+P(B^c)-P\underbrace{(A^c \cap B^c)}_{(A \cup B)^c ~, ~\text{by de Morgan's law}}=(1-P(A))+(1-P(B))-(1-P(A \cup B))$
...