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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? – Simon Hayward Nov 3 '12 at 12:58
Yes. I'm non-native in English. – student Nov 3 '12 at 12:58
That's fine. It's important to be precise of course! :) – Simon Hayward Nov 3 '12 at 13:00
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. – Did Nov 3 '12 at 13:02
@SimonHayward Independent is all right. Independent of each other does not exist. – Did Nov 3 '12 at 13:03
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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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Okay. It is 0.88. – student Nov 3 '12 at 13:06
Don't forget to accept the answer if this has helped you :) – Simon Hayward Nov 3 '12 at 13:17
Sorry, I try to remember. :) – student Nov 3 '12 at 13:32
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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? – Simon Hayward Nov 3 '12 at 13:07
@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. – Dilip Sarwate Nov 3 '12 at 15:04
FWIW I wasn't the one doing the down voting. – Simon Hayward Nov 3 '12 at 15:18

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