Mutually Exclusive Events

• Mar 10th 2012, 08:34 AM
primeimplicant
Mutually Exclusive Events
Hello people, i need some help :)

This is the Question:

Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.

So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)

I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).

• Mar 10th 2012, 08:52 AM
Plato
Re: Mutually Exclusive Events
Quote:

Originally Posted by primeimplicant
This is the Question:
Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.
So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)
I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).

You ought to know that:
$\displaystyle 1\ge\mathcal{P}(A\cup B\cup C)$.
What if $\displaystyle A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?$
• Mar 10th 2012, 01:35 PM
primeimplicant
Re: Mutually Exclusive Events
Quote:

Originally Posted by Plato
You ought to know that:
$\displaystyle 1\ge\mathcal{P}(A\cup B\cup C)$.

Hey Plato, thanks for your response ;-)

I know that, apparently every probabillity must be <=1
Quote:

Originally Posted by Plato
What if $\displaystyle A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?$

Err, that means that the above are mutually exclusive.

But, how am i supposed to prove this nice? What steps should i take ?
• Mar 10th 2012, 01:44 PM
Plato
Re: Mutually Exclusive Events
Quote:

Originally Posted by primeimplicant
that means that the above are mutually exclusive.
But, how am i supposed to prove this nice? What steps should i take ?

This is a basic axiom of probability.
If $\displaystyle \left\{ {{A_n}} \right\}$ is any collection of pair-wise disjoint evednts then
$\displaystyle \mathcal{P}\left( {\bigcup\limits_n {{A_n}} } \right) = \sum\limits_n {\mathcal{P}\left( {{A_n}} \right)} \le 1$
• Mar 10th 2012, 02:10 PM
primeimplicant
Re: Mutually Exclusive Events
Quote:

Originally Posted by Plato
This is a basic axiom of probability.
If $\displaystyle \left\{ {{A_n}} \right\}$ is any collection of pair-wise disjoint evednts then
$\displaystyle \mathcal{P}\left( {\bigcup\limits_n {{A_n}} } \right) = \sum\limits_n {\mathcal{P}\left( {{A_n}} \right)} \le 1$

Ok cool,

After some thought....

$\displaystyle {P}(A\cup B\cup C)$ = P(A)+P(B)+P(C) - P(A/\B) -P(A/\C)-P(B/\C) +P(A/\B/\C)

So, if the events are mutually exclusive, then P(A/\B)=P(A/\C)=P(B/\C)=P(A/\B/\C)$\displaystyle =\emptyset.~$ Is it alright up to this point ?

Then, i conclude that : $\displaystyle {P}(A\cup B\cup C)$ = P(A)+P(B)+P(C) .

BUT P(A)+P(B)+P(C) = 13/12 . So if my calculation is right, P(A)+P(B)+P(C) > 1 implies that $\displaystyle {P}(A\cup B\cup C)$ > 1 . Thats against the basic axiom of probability.

Thus, there is no truth on the given statement.

Its this right, or i am missing something?

Thanks!!
• Mar 10th 2012, 02:15 PM
Plato
Re: Mutually Exclusive Events
Quote:

Originally Posted by primeimplicant
then, i conclude that : $\displaystyle {p}(a\cup b\cup c)$ = p(a)+p(b)+p(c) .
But p(a)+p(b)+p(c) = 13/12 . So if my calculation is right, p(a)+p(b)+p(c) > 1 implies that $\displaystyle {p}(a\cup b\cup c)$ > 1 . Thats against the basic axiom of probability.
Thus, there is no truth on the given statement.

correct!
• Mar 10th 2012, 02:20 PM
primeimplicant
Re: Mutually Exclusive Events
Quote:

Originally Posted by Plato
correct!

Kudos!! :D That was pretty basic i guess, eh ? I am so beginner lol ... Thanks again pal :)