Three machines turn out all the products in a factory, with the first machine producing 35% of the products, the second machine 40%, and the third machine 25%. The first machine produces defective products 2% of the time, the second machine 1% of the time and the third machine 12% of the time. What is the probability that a non-defective product came from the second machine? (Round your answer to four decimal places.)

This is a problem that can be solved using **Bayes' Theorem**, which helps us find the conditional probability of an event given that another event has occurred.

Let’s define the events:
- \( A_1 \): Product came from machine 1
- \( A_2 \): Product came from machine 2
- \( A_3 \): Product came from machine 3
- \( B \): The product is **non-defective**

We are asked to find the probability that a non-defective product came from machine 2, i.e., \( P(A_2 \mid B) \).

### Step 1: Known information
- \( P(A_1) = 0.35 \) (35% of products are from machine 1)
- \( P(A_2) = 0.40 \) (40% of products are from machine 2)
- \( P(A_3) = 0.25 \) (25% of products are from machine 3)
- Defective rates:
  - \( P(\text{Defective} \mid A_1) = 0.02 \)
  - \( P(\text{Defective} \mid A_2) = 0.01 \)
  - \( P(\text{Defective} \mid A_3) = 0.12 \)

Thus, the probabilities of producing a **non-defective** product are:
- \( P(B \mid A_1) = 1 - 0.02 = 0.98 \)
- \( P(B \mid A_2) = 1 - 0.01 = 0.99 \)
- \( P(B \mid A_3) = 1 - 0.12 = 0.88 \)

### Step 2: Use Bayes' Theorem

Bayes' Theorem formula is:
\[
P(A_2 \mid B) = \frac{P(B \mid A_2) P(A_2)}{P(B)}
\]

To find \( P(A_2 \mid B) \), we first need \( P(B) \), the total probability of a non-defective product.

\[
P(B) = P(B \mid A_1) P(A_1) + P(B \mid A_2) P(A_2) + P(B \mid A_3) P(A_3)
\]
Substituting the known values:
\[
P(B) = (0.98)(0.35) + (0.99)(0.40) + (0.88)(0.25)
\]

Now, let’s calculate that:

\[
P(B) = 0.343 + 0.396 + 0.22 = 0.959
\]

### Step 3: Find \( P(A_2 \mid B) \)

Now that we have \( P(B) = 0.959 \), we can substitute it back into Bayes' Theorem:
\[
P(A_2 \mid B) = \frac{(0.99)(0.40)}{0.959} = \frac{0.396}{0.959}
\]

Let's calculate the final result:

\[
P(A_2 \mid B) \approx 0.4129
\]

### Final Answer:
The probability that a non-defective product came from the second machine is approximately **0.4129**.

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the answer change if the defective rates of the machines were different?
2. What is the probability that a defective product came from machine 3?
3. How would you calculate the probability that a product is defective overall?
4. What assumptions are made in using Bayes' Theorem in this problem?
5. Can you extend this problem to more than three machines?

**Tip:** When solving probability problems, especially those involving conditional probability, organizing the given data clearly and methodically applying Bayes' Theorem can greatly simplify the solution process.

Calculate the probability that a non-defective product came from the second machine using Bayes' Theorem. This problem explores key probability concepts, including conditional probability and the application of Bayes' Theorem. Suitable for advanced high school or college-level students.

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