# Probability independence problem

1. Apr 2, 2012

### naptor

1. The problem statement, all variables and given/known data
Let
$E= A \cup \bar{B}$ and $F= \bar{D} \cup C$
Assuming that A,B,C,D are independent show that
F and E are independent

2. Relevant equations
By definition A and by are independent if and only P(AB)=P(A)P(B).
3. The attempt at a solution
I tried to use set theory to simplify E$\cap$F.But I couldn't imply the definition,all I got was this bunch of unions :
EF=$A \bar{D} \cup A C \cup \bar{B} \bar{D} \cup \bar{B} C$

2. Apr 2, 2012

### tiny-tim

hi naptor!

start with, what is $P(A \cup \bar{B})$ ?

3. Apr 2, 2012

### naptor

Hi tim :smile thanks for the reply

ok so :$P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A \cap \bar{B})$ I can get rid off $A \cap \bar{B}$ using $P(A)=P(A\cap B )+P(A \cap \bar{B}) \Rightarrow P(A \cap \bar{B}) =P(A)-P(A\cap B )$ now I can plug this in my first eq and use the hypothesis I got:$P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A)P(\bar{B})$ and the same thing for $P(\bar{D}\cup C)$. I'm trying to combine these expressions in oder to get $P(E \cap F)=P(E)P(F)$.
Am I on the right track?

Last edited: Apr 2, 2012
4. Apr 2, 2012

### tiny-tim

hi naptor!
a bit long-winded, and you still have a P(A), which should have cancelled

easier would have been $P(A \cup \bar{B})=P(\bar{B})+P(A \cap B)$

try again

5. Apr 2, 2012

### Ray Vickson

It is a lot easier to first recognize some easily-proven preliminary properties: (i) two events $U \text{ and } V$ are independent if and only if $U \text{ and } \bar{V}$ are independent; and (ii) If $U, V, W$ are independent, then $U \text{ and } V \cup W$ are independent, as are $U \text{ and } V \cap W$.

Because of these properties it does not matter whether we use $B \text{ or } \bar{B}$ and it does not matter whether we use $C \text{ or } \bar{C}.$ So, we can ask instead whether $A \cup B \text{ and } \bar{C} \cup \bar{D}$ are independent, or equivalently, whether $A \cup B \text{ and } C \cap D$ are independent. This last form is easier to work with. We have
$$P\{ (A \cup B)\cap( C \cap D) \} = P\{ (A \cap C \cap D) \cup (B \cap C \cap \D) \}\\ = P\{A \cap (C \cap D) \} + P\{B \cap (C \cap D) \} – P\{ (A \cap B)\cap(C \cap D)\} \\ = P\{A\} P\{C \cap D\} + P\{B\} P\{C \cap D\} – P\{A \cap B \}P\{ C \cap D\}\\ = P\{ A \cup B\} P\{ C \cap D\},$$
so $A \cup B \text{ and } C \cap D$ are independent.

RGV

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