The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, 4% on Wednesday, 4% on Thursday, 8% on Monday, and 12% on Friday.
The Company manufactures an equal amount of ball bearings (20%) on each weekday. What is the probability that a defective ball bearing was manufactured on a Friday?

To solve the problem, we can apply **Bayes' Theorem**. The goal is to find the probability that a defective ball bearing was manufactured on a Friday.

Let's define the events:
- \( D \): The event that a ball bearing is defective.
- \( F \): The event that a ball bearing was manufactured on Friday.

We need to calculate \( P(F \mid D) \), the probability that the ball bearing was manufactured on Friday given that it is defective.

Using Bayes' Theorem:
\[
P(F \mid D) = \frac{P(D \mid F) \cdot P(F)}{P(D)}
\]

Where:
- \( P(D \mid F) \) is the probability of a ball bearing being defective given it was manufactured on Friday.
- \( P(F) \) is the probability that a ball bearing was manufactured on Friday.
- \( P(D) \) is the total probability that a ball bearing is defective.

### Given Data:
- \( P(D \mid \text{Monday}) = 0.08 \)
- \( P(D \mid \text{Tuesday}) = 0.04 \)
- \( P(D \mid \text{Wednesday}) = 0.04 \)
- \( P(D \mid \text{Thursday}) = 0.04 \)
- \( P(D \mid \text{Friday}) = 0.12 \)
- \( P(\text{Monday}) = P(\text{Tuesday}) = P(\text{Wednesday}) = P(\text{Thursday}) = P(\text{Friday}) = 0.20 \) (since the production is equal across weekdays)

### Step 1: Calculate \( P(D) \)
\( P(D) \) is the total probability that a ball bearing is defective, considering all days:
\[
P(D) = \sum_{i} P(D \mid \text{Day}_i) \cdot P(\text{Day}_i)
\]
Substituting the values:
\[
P(D) = (0.08 \times 0.20) + (0.04 \times 0.20) + (0.04 \times 0.20) + (0.04 \times 0.20) + (0.12 \times 0.20)
\]
\[
P(D) = 0.016 + 0.008 + 0.008 + 0.008 + 0.024 = 0.064
\]

### Step 2: Calculate \( P(F \mid D) \)
Now, we can substitute the values into Bayes' Theorem:
\[
P(F \mid D) = \frac{P(D \mid F) \cdot P(F)}{P(D)} = \frac{0.12 \times 0.20}{0.064}
\]
\[
P(F \mid D) = \frac{0.024}{0.064} = 0.375
\]

### Final Answer:
The probability that a defective ball bearing was manufactured on a Friday is **0.375** or **37.5%**.

Would you like more details on this solution, or do you have any other questions?

Here are 5 related questions you might consider:
1. How would the probability change if more ball bearings were produced on certain days?
2. Can Bayes' Theorem be applied to problems in fields other than manufacturing?
3. What if the probabilities of defects on different days were dependent on each other?
4. How would the probability calculation change if the defect rates were unknown?
5. How can this approach be used to identify the day with the highest probability of defects?

**Tip:** Bayes' Theorem is a powerful tool for updating probabilities based on new information. It is widely used in various fields, including medical diagnostics, finance, and machine learning.

Learn how to calculate the probability that a defective ball bearing was manufactured on a Friday using Bayes' Theorem. Understand conditional probability and apply it to manufacturing data. Suitable for advanced high school or college-level students.

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