scresthop123
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IF A and B are independent events, show A complement (not A) and B complement (not B) are also independent.
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Probability question: please help!
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Hello, scresthop123!
\(\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: }\(A' \cap B') \;=\;P(A')\cdot P(B')\)
\(\displaystyle \text{Therefore, }A'\text{ and }B'\text{ are independent.}\)
\(\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]\(\displaystyle \text{The complement of }A \cup B\text{ is: }\
A \cup B)' \:=\:A' \cap B'\)\(\displaystyle \text{Hence: }\
(A'\cap B')\,\text{ is the "complement" of [1].}\). . \(\displaystyle P(A' \cap B') \;=\;1 - \bigg[P(A) + P(B) - P(A)\cdot P(B)\bigg]\)
. . . . . . . . . . \(\displaystyle =\;1 - P(A) - P(B) + P(A)\cdot P(B)\)
. . . . . . . . . . \(\displaystyle =\; \bigg[1 - P(A)\bigg] \:-\: P(B)\bigg[1 - P(A)\bigg]\)
. . . . . . . . . . \(\displaystyle =\; \bigg[1-P(A)]\cdot\bigg[1-P(B)\bigg]\)
. . . . . . . . . . \(\displaystyle =\qquad P(A')\cdot P(B')\)
\(\displaystyle \text{We have shown that: }\
(A' \cap B') \;=\;P(A')\cdot P(B')\)\(\displaystyle \text{Therefore, }A'\text{ and }B'\text{ are independent.}\)