# Probability proof?

• Sep 28th 2009, 08:09 AM
vexiked
Probability proof?
This is really confusing to me... how could I prove this?

If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that

P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)
• Sep 28th 2009, 10:06 AM
Moo
Hello,
Quote:

Originally Posted by vexiked
This is really confusing to me... how could I prove this?

If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that

P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)

This can be found here : Inclusion-exclusion principle - Wikipedia, the free encyclopedia

For a simple proof... :
P(at least one Ai)= $\mathbb{P}(A_1\cup A_2\cup A_3)$

Now we know that $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$

Apply this to $(A_1\cup A_2)$ and $A_3$

Then try to go from here.
• Sep 30th 2009, 02:21 PM
vexiked
I think I got it with what you have shown but not sure....

P(A1 n A2 n A3) = P(A1) + P(A2) + P(A3) - P(A1 n A2) - P(A1 n A2) + P(A1 n A2 n A3)
• Oct 1st 2009, 06:39 AM
vexiked
Can someone help me out with this one? I'm not sure if I am following the above posters directions correctly.
• Oct 1st 2009, 07:12 AM
Plato
Quote:

Originally Posted by vexiked
Can someone help me out with this one? I'm not sure if I am following the above posters directions correctly.

Quote:

Originally Posted by vexiked
If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that
P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)

.
There very little there that makes any sense what so ever.
The notation is incorrect; the vocabulary is wrong.
Example: “union of A1 and A2 != 0” has no meaning whatsoever.
Neither does “the union of A2 and A3 = 0”.

I suspect that you don’t know union from intersection.

The inclusion/exclusion principle applied to probability function is:
$P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {A \cap C} \right) -$ $P\left( {C \cap B} \right) + P\left( {A \cap B \cap C} \right)$
• Oct 1st 2009, 07:14 AM
vexiked
Oh I am sorry. I definately meant to use intersection.. here it goes with some lateX. Thanks again.

If A1, A2, and A3 are three events. The events A2 and A3 are mutually exclusive.

Show that
$
P\left( at least one Ai \right) = P\left( A1 \right) + P\left( A2\right) + P\left( A3 \right) - 2P\left( {A1 \cap A2} \right)$
• Oct 1st 2009, 07:35 AM
Plato
Quote:

Originally Posted by vexiked
Oh I am sorry. I definately meant to use intersection.. here it goes with some lateX. Thanks again.

If A1, A2, and A3 are three events. The events A2 and A3 are mutually exclusive.

Show that
$
P\left( at least one Ai \right) = P\left( A_1 \right) + P\left( A_2\right) + P\left( A_3 \right) - \color{red}{2P\left( {A_1 \cap A_2} \right)}$

Well that is false unless you are given that:
$P\left( A_1 \cap A_2 \right)=P\left( {A_1 \cap A_3} \right)$ and $A_2\cap A_3=\emptyset$

BTW: Learn to make subscripts.
$$A_2\cap A_3=\emptyset$$ gives $A_2\cap A_3=\emptyset$.
• Oct 1st 2009, 07:55 AM
vexiked
If $A_1, A_2, A_3$ are three events and $P\left(A_1 \cap A_2 \right) = P\left(A_1 \cap A_3 \right) \neq 0$ but $P\left(A_2 \cap A_3\right) = 0$, show that,
$