# independent events

• Sep 27th 2010, 08:00 AM
0123
independent events
hi! I need help understanding these two very close questions.

Let A and B be independent events. Show that the "complement of A" and "B" are independent.
The solution here starts by sayng that the intersection between A-complement and B is equal to B minus the intersection between A and B. But I don't see where this comes from...

The other one is: if P(A)=1 or 0, show that A is independent of all events B.

Thanks!
• Sep 27th 2010, 08:43 AM
HappyJoe
As for the first question, since $A$ and $B$ are independent, we have

$P(A\cap B) = P(A)P(B).$

Let $A^c$ denote the complement of $A$. Then they state that

$A^c\cap B = B\setminus(A\cap B).$

This is correct - for instance, try checking it with a Venn diagram.

Thus:
$P(A^c\cap B) = P(B\setminus(A\cap B))$
$= P(B) - P(A\cap B)$
$= P(B)-P(A)P(B)$
$= P(B)(1-P(A))$
$=P(B)P(A^c),$

which means that $A^c$ and $B$ are independent.

For the other problem: What if $P(A)=0$? Then as $A\cap B$ is a subset of $A$, we have also $P(A\cap B)=0$, which equals $0\cdot P(B) = P(A)\cdot P(B)$.

What if $P(A)=1$? Then the event $A$ definitely occurs, no matter what $B$ is. See if you can write it down formally.
• Sep 27th 2010, 08:46 AM
0123
Why is that $A\cap B$ is a subset of $A$?
• Sep 27th 2010, 09:47 AM
HappyJoe
Quote:

Originally Posted by 0123
Why is that $A\cap B$ is a subset of $A$?

Remember what it means for one set to be a subset of another. The set $C$ is a subset of the set $D$, if and only if all elements of $C$ are also elements of $D$.

So to see that $A\cap B$ is a subset of $A$, you need to convince yourself that all elements of $A\cap B$ are also elements of $A$. In particular, take an element $x$ in $A\cap B$. By definition of intersection, $x$ is an element of $A$ and of $B$. In particular, $x$ is an element of $A$, as we set out to show.
• Sep 27th 2010, 10:32 AM
0123
I still don't get why B shall be contained in A, which is equivalent to say that the intersection is a subset of A...
• Sep 27th 2010, 12:02 PM
MathoMan
if two events A and B are independent then probability of their intersection can be calculated as a product of their probabilities. Contrary to a popular belief, if for two events A and B one shows that the probability of their intersection can be calculated as a product of their probabilities it does not mean that the two events are independent. Sorry it just doesn't work that way.
• Sep 27th 2010, 02:40 PM
Plato
Quote:

Originally Posted by MathoMan
Contrary to a popular belief, if for two events A and B one shows that the probability of their intersection can be calculated as a product of their probabilities it does not mean that the two events are independent. Sorry it just doesn't work that way.

Would you care to provide an example to support the extraordinary claim?
Independence (probability theory) - Wikipedia, the free encyclopedia
• Sep 27th 2010, 03:03 PM
0123
okay.. fiiuu! I was getting really confused at that point..
so now, I still don't get the fact that B is inside A...
• Sep 27th 2010, 03:10 PM
Plato
Quote:

Originally Posted by 0123
okay.. fiiuu! I was getting really confused at that point..
so now, I still don't get the fact that B is inside A...

I have absolutely no idea what that post means.
What are you trying to ask?
Try using standard English.
• Sep 28th 2010, 05:45 AM
HappyJoe
We assume no where that B is inside of A. Maybe you are thinking of something else, like $A\cap B$ being inside of A. But this latter claim is always true, no matter how A and B are positioned with respect to one another, as my argument above shows.
• Sep 28th 2010, 10:08 AM
0123
Okay, but then we can also write $A\cap B$ is inside of B.
I was trying to eason in that way for P(A)=1 but I don't seem to go further...

Quote:

Originally Posted by HappyJoe
We assume no where that B is inside of A. Maybe you are thinking of something else, like $A\cap B$ being inside of A. But this latter claim is always true, no matter how A and B are positioned with respect to one another, as my argument above shows.

• Sep 28th 2010, 10:46 AM
Plato
Quote:

Originally Posted by 0123
Okay, but then we can also write $A\cap B$ is inside of B.
I was trying to eason in that way for P(A)=1 but I don't seem to go further...

If $P(A)=1$ then $1=P(A)\le P(A\cup B)\le 1$ or $P(A\cup B)=1$

Now consider $P(B|A)=\dfrac{P(A\cap B)}{P(A)} =P(A\cap B)=
P(A)+P(B)-P(A\cup B)=P(B).$