Let's break this problem into two parts:

Part 1: What portion of all the parts is defective?

We can solve this part using the law of total probability.

Define the following events:

• $D$: The event that a part is defective.
• $S_1$: The event that a part comes from supplier $S_1$.
• $S_2$: The event that a part comes from supplier $S_2$.
• $S_3$: The event that a part comes from supplier $S_3$.

The problem provides the following information:

• $P(S_1) = 0.50$ (50% of parts come from $S_1$)
• $P(S_2) = 0.20$ (20% of parts come from $S_2$)
• $P(S_3) = 0.30$ (30% of parts come from $S_3$)

Also, the probability that a part is defective given that it came from each supplier:

• $P(D | S_1) = 0.05$ (5% of $S_1$'s parts are defective)
• $P(D | S_2) = 0.03$ (3% of $S_2$'s parts are defective)
• $P(D | S_3) = 0.06$ (6% of $S_3$'s parts are defective)

The total probability that a part is defective can be found using the law of total probability:

$P(D) = P(D | S_1) \cdot P(S_1) + P(D | S_2) \cdot P(S_2) + P(D | S_3) \cdot P(S_3)$

Now, substitute the values:

$P(D) = (0.05 \times 0.50) + (0.03 \times 0.20) + (0.06 \times 0.30)$

$P(D) = 0.025 + 0.006 + 0.018 = 0.049$

Thus, the total portion of defective parts is 4.9%.

Part 2: What is the probability that the defective part was supplied by $S_1$?

For this part, we can use Bayes' Theorem. We want to find $P(S_1 | D)$, the probability that a defective part came from $S_1$, given that the part is defective.

Bayes' Theorem states:

$P(S_1 | D) = \frac{P(D | S_1) \cdot P(S_1)}{P(D)}$

We already know the following:

• $P(D | S_1) = 0.05$
• $P(S_1) = 0.50$
• $P(D) = 0.049$

Now substitute the values:

$P(S_1 | D) = \frac{0.05 \times 0.50}{0.049}$

$P(S_1 | D) = \frac{0.025}{0.049} \approx 0.5102$

Thus, the probability that a defective part was supplied by $S_1$ is approximately 51.02%.

Summary:

1. **The portion of all parts that are