If A and B are independent events, prove A' and B' are also independent events.

[dxoo]

Problem Statement:
If A and B are independent events, prove A' and B' are also independent events.

My Approach:
If A and B are independent:
[MATH] P(A {\displaystyle \cap } B) = P(A) * P(B)\\ P(A'{\displaystyle \cap } B') = P(A')*P(B')\\ P(A{\displaystyle \cup }B)'\\ = 1-P(A{\displaystyle \cup}B)\\ = 1-[P(A) + P(B) - P(A{\displaystyle \cap }B)\\ = 1-P(A)-P(B)-P(A)P(B)\\ = P(A') - P(B) + P(A)P(B)\\ = P(A')-P(B)[1-P(A)]\\ = P(A')-[P(B)P(A')]\\ = P(A')[1-P(B)]\\ = P(A')P(B')\\ [/MATH]
Edit: Looks like I was just missing a few steps, sorry to bother!

 

[Deleted member 4993]

Problem Statement:
If A and B are independent events, prove A' and B' are also independent events.

My Approach:
If A and B are independent:
[MATH] P(A {\displaystyle \cap } B) = P(A) * P(B)\\ P(A'{\displaystyle \cap } B') = P(A')*P(B')\\ P(A{\displaystyle \cup }B)'\\ = 1-P(A{\displaystyle \cup}B)\\ = 1-[P(A) + P(B) - P(A{\displaystyle \cap }B)\\ = 1-P(A)-P(B)-P(A)P(B)\\ = P(A') - P(B) + P(A)P(B)\\ = P(A')-P(B)[1-P(A)]\\ = P(A')-[P(B)P(A')]\\ = P(A')[1-P(B)]\\ = P(A')P(B')\\ [/MATH]
Edit: Looks like I was just missing a few steps, sorry to bother!

Good Job!

 

[pka]

Problem Statement:
If A and B are independent events, prove A' and B' are also independent events.

We know that \(\mathcal{P}(A\cap B')=\mathcal{P}(A)-\mathcal{P}(A\cap B).\)
The from independence we get:
\(\mathcal{P}(A\cap B')=\mathcal{P}(A)-\mathcal{P}(A\cap B)\)
\(=\mathcal{P}(A)-\mathcal{P}(A)\mathcal{P}(B)\)
\(=\mathcal{P}(A)[1-\mathcal{P}(B)]\)
\(=\mathcal{P}(A)[{P}(B')]\)
or \(A~\&~B'\) are independent.
Well if \(A~\&~B'\) are independent so are \(B'~\&~A'\) independent by the same reasoning.