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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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!
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.