This problem can be solved using Bayes' Theorem, which helps in determining conditional probabilities. The event we are interested in is that an item is found defective, and we need to find the probability that this defective item came from Machine I, Machine II, or Machine III.

Let’s define the events:

• $A_1$: The item is produced by Machine I.
• $A_2$: The item is produced by Machine II.
• $A_3$: The item is produced by Machine III.
• $D$: The item is defective.

We are asked to find:

1. $P(A_1 | D)$ — the probability that the defective item came from Machine I.
2. $P(A_2 | D)$ — the probability that the defective item came from Machine II.
3. $P(A_3 | D)$ — the probability that the defective item came from Machine III.

Step 1: Find the prior probabilities

The daily production by each machine gives us the prior probabilities $P(A_1)$, $P(A_2)$, and $P(A_3)$.

• Total production = $3000 + 2500 + 4500 = 10000$ units.
• $P(A_1) = \frac{3000}{10000} = 0.3$
• $P(A_2) = \frac{2500}{10000} = 0.25$
• $P(A_3) = \frac{4500}{10000} = 0.45$

Step 2: Find the likelihoods

The likelihoods represent the probability of drawing a defective item given that it came from a specific machine. These are provided in the problem:

• $P(D | A_1) = 0.01$ (1% defective from Machine I)
• $P(D | A_2) = 0.012$ (1.2% defective from Machine II)
• $P(D | A_3) = 0.02$ (2% defective from Machine III)

Step 3: Apply the Law of Total Probability

We first need to find the overall probability of drawing a defective item, $P(D)$, by using the law of total probability:

$P(D) = P(D | A_1)P(A_1) + P(D | A_2)P(A_2) + P(D | A_3)P(A_3)$

Substituting the known values:

$P(D) = (0.01)(0.3) + (0.012)(0.25) + (0.02)(0.45)$

$P(D) = 0.003 + 0.003 + 0.009 = 0.015$

Step 4: Use Bayes' Theorem

Now we can apply Bayes' Theorem to calculate the conditional probabilities:

$P(A_i | D) = \frac{P(D | A_i) P(A_i)}{P(D)}$

(a) Probability that the defective item came from Machine I:

$P(A_1 | D) = \frac{(0.01)(0.3)}{0.015} = \frac{0.003}{0.015} = 0.2$

(b) Probability that the defective item came from Machine II:

$P(A_2 | D) = \frac{(0.012)(0.25)}{0.015} = \frac{0.003}{0.015} = 0.2$

(c) Probability that the defective item came from Machine III:

$P(A_3 | D) = \frac{(0.02)(0.45)}{0.015} = \frac{0.009}{0.015} = 0.6$

Final Answer:

• $P(A_1 | D) = 0.2$
• $P(A_2 | D) = 0.2$
• $P(A_3 | D) = 0.6$

Thus, if the item is found to be defective:

• There is a 20% chance it came from Machine I,
• A 20% chance it came from Machine II, and
• A 60% chance it came from Machine III.

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Would you like further clarification on any of the steps or calculations? Here are some related questions to explore:

1. How does Bayes' Theorem apply to real-life quality control processes?
2. How would the result change if Machine III produced 5% defective items instead of 2%?
3. What happens if all machines produced the same number of units but had different defect rates?
4. How can we extend this method to more than three machines with similar conditions?
5. What would be the effect of increasing the overall production while keeping defect percentages constant?

Tip: Bayes' Theorem is particularly useful when we need to update our belief about a cause (which machine produced a defective item) based on observed evidence (the defect).