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If we have an event $A$ and a sample space $\Omega$, can we say that the event $A$ is Independent on an event $B$ if the occurrence of $B$ keep the ratio of $\frac{|A|}{|\Omega|}$?

For example: looking at a deck of cards, P(heart)=$\frac{13}{52}$ and P(king)=$\frac{4}{52}$ assume we show the card is red.

now P(heart|red)=$\frac{13}{52-26}=\frac{13}{26}\neq$ P(heart)=$\frac{13}{52}$

but P(king|red)=$\frac{4-2}{52-26}=\frac{2}{26}=$P(king)=$\frac{4}{52}=\frac{2}{26}$

gbox
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  • I think you have it right, but the phrasing is a little unclear. The usual definition is that P(A|B) = P(A), i.e. "the probability that A has occurred is unchanged if we are told that B has occurred". – lulu Jul 08 '15 at 17:10
  • @lulu Independence is defined as $P(A \cap B) = P(A) P(B)$ and not in terms of conditional probability. The two definitions are not equivalent because conditional probability is undefined when the conditioning event has probability zero. – dsaxton Jul 08 '15 at 21:01

1 Answers1

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Yes, by Definition you have for the conditional probability that, given $P(B)>0$, $P(A|B) = \frac {P(A \cap B)}{P(B)} $. Therefore if A and B are Independent, we have $P(A \cap B) = P(A) * P(B)$. Then $P(A|B) = P(A)$, so indeed the probability remains unchanged.

Fluffy12
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