# show that events are independent

• Jan 11th 2009, 01:03 PM
Yan
show that events are independent
Show that if events A and B are independent, then events A' and B are independent.
• Jan 11th 2009, 04:56 PM
Last_Singularity
Quote:

Originally Posted by Yan
Show that if events A and B are independent, then events A' and B are independent.

If events $\displaystyle A,B$ are independent, then we know that
$\displaystyle P(A \cap B) = P(A) P(B)$ by definition. That's our key.

For the proof, we will use the following lemma:
$\displaystyle A^C \cap B = B \setminus (A \cap B)$

Proof of the lemma:
$\displaystyle A^C \cap B = (\Omega \setminus A) \cap B = (\Omega \cap B) \setminus (A \cap B) = B \setminus (A \cap B)$

So now we know that lemma is true, we can use it to conclude that:
$\displaystyle P(A^C \cap B) = P(B) - P(A \cap B)$

Looking at the right hand side, we have:
$\displaystyle =P(B) - P(A \cap B)$
$\displaystyle =P(B) - P(A)P(B)$ (using our key assumption)
$\displaystyle =(1-P(A))P(B)$
$\displaystyle =P(A^C)P(B)$

So we conclude that
$\displaystyle P(A^C \cap B) = P(A^C)P(B)$ as desired.

$\displaystyle QED$
• Jan 11th 2009, 05:31 PM
Plato
$\displaystyle \begin{array}{*{20}c} {P\left( {A^c \cap B} \right)} & = & {P\left( B \right) + P\left( {A \cap B} \right)} \\ {} & = & {P\left( B \right) + P\left( A \right)P\left( B \right)} \\ {} & = & {P(B)\left( {1 - P(A)} \right)} \\ {} & = & {P(B)P\left( {A^c } \right)} \\ \end{array}$
• Jan 11th 2009, 05:35 PM
mr fantastic
Quote:

Originally Posted by Plato
$\displaystyle \begin{array}{*{20}c} {P\left( {A^c \cap B} \right)} & = & {P\left( B \right) {\color{red} - } P\left( {A \cap B} \right)} \\ {} & = & {P\left( B \right) {\color{red} - } P\left( A \right)P\left( B \right)} \\ {} & = & {P(B)\left( {1 - P(A)} \right)} \\ {} & = & {P(B)P\left( {A^c } \right)} \\ \end{array}$

Small typos fixed (in red).

A Karnaugh table is a visual way of making both the first line of Plato's proof and the conclusion of Last Singularity's lemma crystal clear.