1. **Problem statement:** Find the probability that an item has defect D2 given it has defect D1. 2. **Explanation:** The conditional probability formula is $$P(D2|D1) = \frac{P(D1 \cap D2)}{P(D1)}$$ where $P(D1 \cap D2)$ is the probability that the item has both defects D1 and D2, and $P(D1)$ is the probability that the item has defect D1. 3. Since the problem does not provide explicit probabilities, we assume the answer choices represent possible values of $P(D2|D1)$. 4. Therefore, the answer is the value among the options that correctly represents $P(D2|D1)$ based on given data (not provided here). --- 1. **Problem statement:** Solve the system of equations: $$\begin{cases} x + y + 2z = 1 \\ x + 2y + z = 1 \\ 3x + 4y + 5z = 7 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 1 & 2 & 1 & 1 \\ 3 & 4 & 5 & 7 \end{array}\right]$$ 3. **Perform row operations:** - Subtract row 1 from row 2: $$R_2 \to R_2 - R_1: \left[\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 0 & 1 & -1 & 0 \\ 3 & 4 & 5 & 7 \end{array}\right]$$ - Subtract 3 times row 1 from row 3: $$R_3 \to R_3 - 3R_1: \left[\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 4 \end{array}\right]$$ - Subtract row 2 from row 3: $$R_3 \to R_3 - R_2: \left[\begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 4 \end{array}\right]$$ 4. The last row corresponds to the equation: $$0x + 0y + 0z = 4$$ which is impossible. 5. **Conclusion:** The system is inconsistent and has no solution. **Answer:** D. $\emptyset$