# Thread: [SOLVED] Mutually Indpendent Question

1. ## [SOLVED] Mutually Indpendent Question

Hi,

I just needed a bit of help with the following question.

Show that if A, B and C are mutually independent, then A and (B U C) are independent.

From what I understand if A, B and C are mutually independent then

P(AnBnC) = P(A)P(B)P(C)
P(AnB) = P(A)P(B)
P(AnC) = P(A)P(C)
P(BnC) = P(B)P(C)

To prove A and (B U C) are independent do I have to prove P(A|(BUC) ) = P(A) because I don't know what to do after
P((BUC)nA)/P(BUC) = P[(BnA)U(AnC)]/P(BUC)

THANK YOU

2. Your definition of "mutual independence" is indeed correct. As you can see, it involves intersections. So you must reduce $P(A\cap(B\cup C))$ to an expression using intersections. The usual trick is to use the complement since $(B\cup C)^c=B^c\cap C^c$ (where $B^c=\Omega\setminus B$ if $\Omega$ is the whole probability space).

So here are the first steps (that you should justify): $P(A\cap(B\cup C))=P(A)-P(A\cap(B\cup C)^c)=P(A)-P(A\cap B^c \cap C^c)$.

You will need to know that if $A, B, C$ are mutually independent, then $A, B^c, C^c$ are mutually independent as well. Either you know it already or you'll have to prove it.

Laurent.

3. Can I continue like this

$

P(A)-P(A\cap B^c \cap C^c) = P(A)-P(A)P( B^c \cap C^c) = P(A)[1 -P( B^c \cap C^c)]

$

$

= P(A)(B\cup C)

$

Therefore they are independent

4. I'd rather insert an intermediate equality $P(A)-P(A)P(B^c)P(C^c)=P(A)-P(A)P(B^c\cap C^c)$ in what you wrote, since you don't know directly that $P(A\cap B^c\cap C^c)=P(A)P(B^c\cap C^c)$ (this is almost what you were asked to prove!). Then this is OK.

5. $\begin{array}{rcl}
{P\left[ {A(B \cup C)} \right]} & = & {P\left( {AB \cup AC} \right)} \\
{} & = & {P(AB) + P(AC) - P(ABC)} \\
{} & = & {P(A)P(B) + P(A)P(C) - P(A)P(B)P(C)} \\
{} & = & {P(A)\left[ {P(B) + P(C) - P(BC)} \right]} \\
{} & = & {P(A)\left[ {P(B \cup C)} \right]} \\ \end{array}$

6. Thanks for all the help!!!

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