1. **Problem statement:** Given two independent events $A$ and $B$ in sample space $\Omega$ with $P(A \cap B) = 0.08$ and $P(A) = 2P(B)$, find $P(A \cup B)$. 2. **Formula used:** For any two events $A$ and $B$, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$. 3. **Step-by-step solution:** - Let $P(B) = p$. - Then $P(A) = 2p$. - Given $P(A \cap B) = 0.08$, but also $P(A \cap B) = P(A)P(B) = 2p \times p = 2p^2$. - So, $2p^2 = 0.08 \Rightarrow p^2 = 0.04 \Rightarrow p = 0.2$ (since probability is positive). - Then $P(A) = 2 \times 0.2 = 0.4$. - Now, calculate $P(A \cup B)$: $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.2 - 0.08 = 0.52$$ **Final answer:** $$P(A \cup B) = 0.52$$