Probability independence problem

[naptor]

Homework Statement

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

Homework Equations

By definition A and by are independent if and only P(AB)=P(A)P(B).

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=$A \bar{D} \cup A C \cup \bar{B} \bar{D} \cup \bar{B} C$

 

[tiny-tim]

hi naptor!

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

 

[naptor]

hi naptor!

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

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?

 

[tiny-tim]

hi naptor!

$P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A \cap \bar{B})$ 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:$P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A)P(\bar{B})$

a bit long-winded, and you still have a P(A), which should have canceled

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

try again ​

 

[Ray Vickson]

Homework Statement

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

Homework Equations

By definition A and by are independent if and only P(AB)=P(A)P(B).

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=$A \bar{D} \cup A C \cup \bar{B} \bar{D} \cup \bar{B} C$

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