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Events $A$ and $B$ are independent. We know their probabilities, $P(A)=0.7, P(B)=0.6$. Compute $P(A \cup B)$? Can this be solved somehow?

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    Do you mean they are independent of each other?2012-11-03
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    Yes. I'm non-native in English.2012-11-03
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    That's fine. It's important to be precise of course! :)2012-11-03
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    Hint: P(A or B) = P(A) + P(B) - P(A and B). (2.) Since A and B are independent, one knows P(A and B). Ergo.2012-11-03
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    @SimonHayward *Independent* is all right. *Independent of each other* does not exist.2012-11-03
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    Yes, should say independent. Original post said A and B do not depend on each other, which I was attempting to correct (badly).2012-11-03
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    @student While we are dealing with terminology questions, please note that what you call *consequences* are really *events*.2012-11-03
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    @did Thanks. I try to remember that next time.2012-11-03

2 Answers 2

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Ok, so since $A$ and $B$ are independent we have $P(A \cap B)= P(A).P(B)$. Further, basic algebra of sets tells us that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

You should be able to take it from here!

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    Okay. It is 0.88.2012-11-03
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    Sorry, I try to remember. :)2012-11-03
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If two events, A and B are independent then the joint probability is P(A AND B) = P(A)XP(B) = 0.7X0.6 = 0.42

fOR, P(A OR B) = P(A) + P(B) - P(A AND B) = 0.7+0.6-0.42 = 0.88.

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    Why give the answer away completely? That's not hugely helpful for learning..... Also, do you know Latex formatting?2012-11-03
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    @SimonHayward Let's cut Maitreyi some slack. This is a first answer on math.SE from someone who has asked seven questions of the kind typically asked by a beginning student of probability. I am glad that Maitreyi has learned something useful and is willing to contribute this knowledge to this forum, elementary though this calculation is.2012-11-03
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    FWIW I wasn't the one doing the down voting.2012-11-03