Question :

Show that if the event A is completely independent of events B and C, then A is independent of logical sum $\displaystyle B \cup C$

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- Dec 2nd 2009, 04:38 PMzorroShow event is completly independent of another event
Question :

Show that if the event A is completely independent of events B and C, then A is independent of logical sum $\displaystyle B \cup C$ - Dec 2nd 2009, 09:56 PMmatheagle
assume that

$\displaystyle P(AB)=P(A)P(B)$ and $\displaystyle P(AC)=P(A)P(C)$

the question is, does

$\displaystyle P(A(B\cup C))=P(A)P(B\cup C)$????????

I don't think so.

Try tossing a fair die.

Let $\displaystyle A=2,4,6$

$\displaystyle B=2,3$

and

$\displaystyle C=2,5$

I get 1/6 =1/6 for the first two

but then I get 1/6 and 1/4 on the last one. - Dec 13th 2009, 12:51 AMzorroCould u please explain me
- Dec 13th 2009, 03:19 AMLaurent
I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that $\displaystyle P(A\cap B)=P(A)P(B)$ (same with A,C and B,C),

**and**$\displaystyle P(A\cap B\cap C)=P(A)P(B)P(C)$.

Then the property is proved as follows (for instance):

$\displaystyle P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c))$, and the three (main) events on the right-hand side are disjoints, hence

$\displaystyle P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c)$,

and you can compute the latter probabilities using $\displaystyle A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C)$ (and similarly for the last term) and the assumptions (recalled above).

You should get $\displaystyle \cdots=P(A)P(B\cup C)$, or in fact $\displaystyle P(A)(P(B)+P(C)-P(B)P(C))$, which is the same... - Dec 15th 2009, 03:42 PMzorroI am again stuck

http://www.mathhelpforum.com/math-he...98af3ab8-1.gif,

$\displaystyle

Pr(A \cap(B \cup C)) = Pr(A \cap B \cap C) + \frac{Pr(A \cap C)}{Pr(A \cap B \cap C)} + \frac{Pr(A \cap B)}{Pr(A \cap B \cap C)}$

I am stuck here ???? - Dec 18th 2009, 01:21 PMLaurent
- Dec 18th 2009, 01:27 PMzorroI did get u