1. ## [solved] p(a)=1

Hi

A little question...

Let $\Omega$ be the probability space.
We know that if $P(A)=1$, then $P(\Omega \backslash A)=0$ (we don't necessarily have $A=\Omega$)

Is there a proof (with formulae, not diagrams) for that :
For any $X \subset \Omega$, if P(A)=1, then $P(A \cap X)=P(X)$
?

Thanks

2. Since

$P(A \cap X)=P(X \mid A) P(A)$
if P(A)=1 its pretty clear right?

3. Originally Posted by Moo
Hi

A little question...

Let $\Omega$ be the probability space.
We know that if $P(A)=1$, then $P(\Omega \backslash A)=0$ (we don't necessarily have $A=\Omega$)

Is there a proof (with formulae, not diagrams) for that :
For any $X \subset \Omega$, if P(A)=1, then $P(A \cap X)=P(X)$
?

Thanks
Here's my proof:
$P(X)=P(A\cap X)+P(A^c\cap X)$ where $A^c=\Omega\setminus A$,
and $0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0$,
so that $P(X)=P(A\cap X)$.

It has the significant advantage (compared to the previous one) of avoiding to divide by 0.

4. Originally Posted by Robert Hall
Since

$P(A \cap X)=P(X \mid A) P(A)$
if P(A)=1 its pretty clear right?
Hmm So how would you prove that $P(X \mid A)=P(X)$ ? That's the problem...

Originally Posted by Laurent
Here's my proof:
$P(X)=P(A\cap X)+P(A^c\cap X)$ where $A^c=\Omega\setminus A$,
and $0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0$,
so that $P(X)=P(A\cap X)$.

It has the significant advantage (compared to the previous one) of avoiding to divide by 0.
That's... erm perfect... :P

Too bad Plato deleted his post, I can't remember it completely ><

Thank you to all three for replying
And special thanks to Laurent, because he can always provide the answer I'm looking for