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\capF.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\capF.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) \}\\<br /> = P\{A \cap (C \cap D) \} + P\{B \cap (C \cap D) \} – P\{ (A \cap B)\cap(C \cap D)\} \\<br /> = P\{A\} P\{C \cap D\} + P\{B\} P\{C \cap D\} – P\{A \cap B \}P\{ C \cap D\}\\<br /> = P\{ A \cup B\} P\{ C \cap D\},
so A \cup B \text{ and } C \cap D are independent.

RGV