Given probabilities of production share:

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):
$P(D|M_1) = 1\% = 0.01$
$P(D|M_2) = 3\% = 0.03$
$P(D|M_3) = 5\% = 0.05$

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 $M_3$: $P(D \cap M_3) = P(D|M_3)P(M_3) = (0.05)(0.50) = 0.025$

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:

The denominator $P(D)$ is the total probability of an item being defective, which is the sum of the probabilities calculated in Step 2:

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.

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 $\boxed{\text{D}}$