# Probabilities, computing when at least one thing happen

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

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