Probability question: please help!

[scresthop123]

IF A and B are independent events, show A complement (not A) and B complement (not B) are also independent.

Thanks.

 

[soroban]

Hello, scresthop123!

If $\displaystyle A$ and $\displaystyle B$ are independent events,
show that: $\displaystyle A'$ and $\displaystyle B'$ are also independent.

\(\displaystyle \text{Since }A\text{ and }B\text{ are independent: }\(A \cap B) \:=\(A)\cdot P(B)\)


$\displaystyle \text{Formula: }\;P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)$

\(\displaystyle \text{We have: }\(A \cup B) \;=\;P(A) + P(B) - P(A)\cdot P(B)\) .[1]


\(\displaystyle \text{The complement of }A \cup B\text{ is: }\A \cup B)' \:=\:A' \cap B'\)

\(\displaystyle \text{Hence: }\(A'\cap B')\,\text{ is the "complement" of [1].}\)

. . $\displaystyle P(A' \cap B') \;=\;1 - \bigg[P(A) + P(B) - P(A)\cdot P(B)\bigg]$

. . . . . . . . . . $\displaystyle =\;1 - P(A) - P(B) + P(A)\cdot P(B)$

. . . . . . . . . . $\displaystyle =\; \bigg[1 - P(A)\bigg] \:-\: P(B)\bigg[1 - P(A)\bigg]$

. . . . . . . . . . $\displaystyle =\; \bigg[1-P(A)]\cdot\bigg[1-P(B)\bigg]$

. . . . . . . . . . $\displaystyle =\qquad P(A')\cdot P(B')$


\(\displaystyle \text{We have shown that: }\(A' \cap B') \;=\;P(A')\cdot P(B')\)

$\displaystyle \text{Therefore, }A'\text{ and }B'\text{ are independent.}$