Kelvin is a manager of an automation design company. Previously, he has 8 engineering staffs reporting to him. Based on his assessment on the 8 engineers, he is only satisfied with the performance of 6 of his engineers. He further concluded from past 5 years observation that if a project was assigned to those engineers who had met his work expectation, 85% of the time the project will be completed before dateline. On the other hand, the 2 engineers whose performance is unsatisfactory has only 35% of the time completed their projects before dateline.

One of the 2 “unsatisfactory” engineers had resigned from the company 8 months ago, and Jimmy had hired 6 months ago.

During that time in the company, Jimmy had successfully completed a project before the given dateline. Based on this information, what is the chance that Jimmy’s performance is meeting Kelvin expectation?

To find the probability that Jimmy meets Kelvin's expectations based on the information given, we can use Bayes' theorem. 

### Step 1: Define the Events

Let:
- \( S \): Jimmy meets Kelvin's expectations.
- \( F \): Jimmy does not meet Kelvin's expectations.
- \( C \): Jimmy completed a project before the deadline.

We need to find \( P(S|C) \), the probability that Jimmy meets Kelvin's expectations given that he completed a project before the deadline.

### Step 2: Identify Probabilities

From the problem:
- Probability of meeting expectations: \( P(S) = \frac{6}{8} = 0.75 \).
- Probability of not meeting expectations: \( P(F) = \frac{2}{8} = 0.25 \).
- Probability of completing the project on time if meeting expectations: \( P(C|S) = 0.85 \).
- Probability of completing the project on time if not meeting expectations: \( P(C|F) = 0.35 \).

### Step 3: Use Bayes' Theorem

Bayes' theorem states:

\[
P(S|C) = \frac{P(C|S) \times P(S)}{P(C)}
\]

where \( P(C) \) is the total probability of completing the project on time, given by:

\[
P(C) = P(C|S) \times P(S) + P(C|F) \times P(F)
\]

Substituting the values:

\[
P(C) = (0.85 \times 0.75) + (0.35 \times 0.25) = 0.6375 + 0.0875 = 0.725
\]

Now, substitute \( P(C) \) into Bayes' theorem:

\[
P(S|C) = \frac{0.85 \times 0.75}{0.725} = \frac{0.6375}{0.725} \approx 0.8793
\]

### Conclusion

The probability that Jimmy’s performance meets Kelvin's expectations, given that he successfully completed a project before the deadline, is approximately **0.8793** or **87.93%**.

---

Would you like more details or have any questions about the calculations? Here are some related questions:

1. How would the probability change if Jimmy completed two projects on time?
2. What if the probability of on-time completion for unsatisfactory engineers was 50%?
3. How does changing the number of satisfactory engineers affect this probability?
4. What are the implications of hiring decisions based on probability?
5. How would this model adjust if engineers' performances were re-evaluated?

**Tip:** Bayes' theorem is particularly useful when updating the probability of an event based on new evidence.

Kelvin is a manager of an automation design company. Previously, he has 8 engineering staff reporting to him. Based on his assessment on the 8 engineers, he is only satisfied with the performance of 6 of his engineers. He further concluded from past 5 years observation that if a project was assigned to those engineers who had met his work expectation, 85% of the time the project will be completed before deadline. On the other hand, the 2 engineers whose performance is unsatisfactory have only 35% of the time completed their projects before the deadline.

One of the 2 'unsatisfactory' engineers had resigned from the company 8 months ago, and Jimmy was hired 6 months ago.

During that time in the company, Jimmy successfully completed a project before the given deadline. Based on this information, what is the chance that Jimmy’s performance is meeting Kelvin's expectation?

Explore how to calculate the probability that a new hire, Jimmy, meets expectations using Bayes' Theorem. This problem involves probability concepts like conditional probability and prior knowledge from performance data. Perfect for advanced high school students or undergraduates studying probability theory.

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