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?
Probabilities, computing when at least one thing happen
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probability
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0Do you mean they are independent of each other? – 2012-11-03
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0Yes. I'm non-native in English. – 2012-11-03
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0That's fine. It's important to be precise of course! :) – 2012-11-03
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0Hint: 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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0@SimonHayward *Independent* is all right. *Independent of each other* does not exist. – 2012-11-03
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0Yes, 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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0@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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0@did Thanks. I try to remember that next time. – 2012-11-03
2 Answers
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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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1Okay. It is 0.88. – 2012-11-03
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0Sorry, 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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1Why 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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0@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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0FWIW I wasn't the one doing the down voting. – 2012-11-03