Question:

Step 1: Determine the total number of articles in Lot C. Lot C is formed by taking 3 articles from Lot A and 4 articles from Lot B. Therefore, the total number of articles in Lot C is: $$\text{Total articles in C} = 3 + 4 = 7$$ Step 2: Calculate the expected number of defective articles from Lot A. Lot A contains 5 good and 3 defective articles, for a total of 8 articles. When we select 3 articles from Lot A, the probability that any single article is defective is $\frac{3}{8}$. The expected number of defective articles from Lot A is: $$E(\text{Defectives from A}) = 3 \times \frac{3}{8} = \frac{9}{8}$$ Step 3: Calculate the expected number of defective articles from Lot B. Lot B contains 3 good and 5 defective articles, for a total of 8 articles. When we select 4 articles from Lot B, the probability that any single article is defective is $\frac{5}{8}$. The expected number of defective articles from Lot B is: $$E(\text{Defectives from B}) = 4 \times \frac{5}{8} = \frac{20}{8} = \frac{5}{2}$$ Step 4: Calculate the total expected number of defective articles in Lot C. The total expected number of defective articles in Lot C is the sum of defectives from both lots: $$E(\text{Defectives in C}) = \frac{9}{8} + \frac{20}{8} = \frac{29}{8}$$ Step 5: Calculate the probability that a randomly chosen article from C is defective. The probability that an article chosen at random from Lot C is defective equals the ratio of the expected number of defective articles to the total number of articles in C: $$P(\text{Defective}) = \frac{E(\text{Defectives in C})}{\text{Total articles in C}} = \frac{\frac{29}{8}}{7} = \frac{29}{8} \times \frac{1}{7} = \frac{29}{56}$$ **Final Answer:** The probability that an article chosen at random from Lot C is defective is $\boxed{\frac{29}{56}}$, which corresponds to **Option 3**. <div class="key-concept"><strong>Key Concept:</strong> Probability using expectation / conditional probability with combinations</div> <div class="trap-box"><strong>Trap:</strong> Students might try to enumerate all cases using combinations instead of using the simpler expectation approach, leading to computational errors.</div>