1. **State the problem:** We are given probabilities $P(A_1)=0.58$, $P(A_2)=0.34$, and $P(A_1 \cap A_2)=0.09$. We want to find the probability that at least one of the two diagnoses detects the defect, i.e., $P(A_1 \cup A_2)$. 2. **Formula used:** The probability of the union of two events is given by the formula: $$ P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) $$ This formula accounts for the overlap counted twice when adding $P(A_1)$ and $P(A_2)$. 3. **Substitute the values:** $$ P(A_1 \cup A_2) = 0.58 + 0.34 - 0.09 $$ 4. **Calculate:** $$ P(A_1 \cup A_2) = 0.92 - 0.09 = 0.83 $$ 5. **Interpretation:** The probability that at least one of the two diagnoses detects the defect is $0.83$, or 83%. This means there is a high chance that the defect will be detected by at least one diagnosis.