A manufacturer has three machine operators A,B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Explanation

To find the probability that a defective item was produced by operator A, we can use Bayes' theorem. Let's break down the solution step by step.

### Step 1: Define Events
Let:
- \( E_1 \): Event that the item was produced by operator A.
- \( E_2 \): Event that the item was produced by operator B.
- \( E_3 \): Event that the item was produced by operator C.
- \( X \): Event that the item is defective.

### Step 2: Determine Probabilities of Events
From the problem statement:
- Probability that an item is produced by A: \( P(E_1) = 0.5 \)
- Probability that an item is produced by B: \( P(E_2) = 0.3 \)
- Probability that an item is produced by C: \( P(E_3) = 0.2 \)

### Step 3: Determine Conditional Probabilities of Defective Items
- Probability that an item produced by A is defective: \( P(X|E_1) = 0.01 \) (1%)
- Probability that an item produced by B is defective: \( P(X|E_2) = 0.05 \) (5%)
- Probability that an item produced by C is defective: \( P(X|E_3) = 0.07 \) (7%)

### Step 4: Calculate Total Probability of Defective Items
Using the law of total probability, we calculate \( P(X) \):
\[ P(X) = P(X|E_1) \cdot P(E_1) + P(X|E_2) \cdot P(E_2) + P(X|E_3) \cdot P(E_3) \]
Substituting the values:
\[ P(X) = (0.01 \cdot 0.5) + (0.05 \cdot 0.3) + (0.07 \cdot 0.2) \]
Calculating each term:
\[ P(X) = 0.005 + 0.015 + 0.014 = 0.034 \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(E_1|X) \), the probability that a defective item was produced by A:
\[ P(E_1|X) = \frac{P(X|E_1) \cdot P(E_1)}{P(X)} \]
Substituting the known values:
\[ P(E_1|X) = \frac{0.01 \cdot 0.5}{0.034} \]
Calculating the numerator:
\[ P(E_1|X) = \frac{0.005}{0.034} \]
Calculating this gives:
\[ P(E_1|X) \approx 0.1471 \]

### Final Answer
Thus, the probability that a defective item was produced by operator A is approximately:
\[ \frac{5}{34} \text{ or } 0.1471 \]