Step 1: Define the events and probabilities. Let $M_1, M_2, M_3$ be the events that an item is produced by machine 1, 2, or 3, respectively. Let $D$ be the event that an item is defective. Given probabilities of production share: $P(M_1) = 20\% = 0.20$ $P(M_2) = 30\% = 0.30$ $P(M_3) = 50\% = 0.50$ Given defect rates (conditional probabilities): $P(D|M_1) = 1\% = 0.01$ $P(D|M_2) = 3\% = 0.03$ $P(D|M_3) = 5\% = 0.05$ Step 2: Calculate the probability of an item being defective from each machine. We need to find $P(D \cap M_i) = P(D|M_i)P(M_i)$ for each machine. This represents the proportion of all items that are defective and come from a specific machine. For $M_1$: $P(D \cap M_1) = P(D|M_1)P(M_1) = (0.01)(0.20) = 0.002$ For $M_2$: $P(D \cap M_2) = P(D|M_2)P(M_2) = (0.03)(0.30) = 0.009$ For $M_3$: $P(D \cap M_3) = P(D|M_3)P(M_3) = (0.05)(0.50) = 0.025$ Step 3: Determine which machine is most likely to have produced a defective item. We are looking for $P(M_i|D)$, which is the probability that a defective item came from machine $M_i$. According to Bayes' Theorem: $$P(M_i|D) = \frac{P(D|M_i)P(M_i)}{P(D)}$$ The denominator $P(D)$ is the total probability of an item being defective, which is the sum of the probabilities calculated in Step 2: $$P(D) = P(D \cap M_1) + P(D \cap M_2) + P(D \cap M_3) = 0.002 + 0.009 + 0.025 = 0.036$$ To find which machine is most likely, we need to compare $P(M_1|D)$, $P(M_2|D)$, and $P(M_3|D)$. Since $P(D)$ is the same for all of them, the machine with the largest $P(D \cap M_i)$ will have the largest $P(M_i|D)$. Comparing the values from Step 2: $P(D \cap M_1) = 0.002$ $P(D \cap M_2) = 0.009$ $P(D \cap M_3) = 0.025$ The largest value is $0.025$, which corresponds to machine $M_3$. Therefore, machine $M_3$ is most likely to have produced the defective item. Let's calculate the exact probabilities for completeness: $P(M_1|D) = \frac{0.002}{0.036} = \frac{2}{36} \approx 0.0556$ $P(M_2|D) = \frac{0.009}{0.036} = \frac{9}{36} = 0.25$ $P(M_3|D) = \frac{0.025}{0.036} = \frac{25}{36} \approx 0.6944$ Machine $M_3$ has the highest probability. Step 4: Evaluate the given options. A. All machines are equally likely, because the item is already known to be defective. (Incorrect) B. $M_1$, because it has the lowest defect rate. (Incorrect, $M_1$ is the least likely) C. $M_2$, because its defect rate is in the middle. (Incorrect) D. $M_3$, because it has the highest defect rate and also the largest production share. (Correct, $M_3$ has $P(D|M_3) = 0.05$ (highest) and $P(M_3) = 0.50$ (largest production share), which together lead to the highest probability of producing a defective item.) The final answer is $\boxed{\text{D}}$ That's 2 down. 3 left today — send the next one.