In a bulb factory, three machines, A, B, C, manufacture 90%, 25% and 15% of the total production respectively. Of their respective outputs, 1 %, 2% and 1 % are defective. A bulb is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by machine C.

Answer

To solve the problem, we will use Bayes' Theorem. Let's break down the steps involved in finding the probability that a defective bulb was manufactured by machine C.

### Step 1: Define the Events
Let:
- \( E_1 \): Event that the bulb is produced by machine A.
- \( E_2 \): Event that the bulb is produced by machine B.
- \( E_3 \): Event that the bulb is produced by machine C.
- \( D \): Event that the bulb is defective.

### Step 2: Determine the Probabilities of Each Event
From the problem, we know:
- \( P(E_1) = 0.90 \) (90% of bulbs are produced by machine A)
- \( P(E_2) = 0.25 \) (25% of bulbs are produced by machine B)
- \( P(E_3) = 0.15 \) (15% of bulbs are produced by machine C)

### Step 3: Determine the Probability of Defectiveness for Each Machine
The probabilities of a bulb being defective given the machine are:
- \( P(D | E_1) = 0.01 \) (1% of A's output is defective)
- \( P(D | E_2) = 0.02 \) (2% of B's output is defective)
- \( P(D | E_3) = 0.01 \) (1% of C's output is defective)

### Step 4: Calculate the Total Probability of a Defective Bulb
Using the law of total probability, we can find \( P(D) \):
\[ P(D) = P(D | E_1)P(E_1) + P(D | E_2)P(E_2) + P(D | E_3)P(E_3) \]
Substituting the values:
\[ P(D) = (0.01)(0.90) + (0.02)(0.25) + (0.01)(0.15) \]
Calculating each term:
- \( (0.01)(0.90) = 0.009 \)
- \( (0.02)(0.25) = 0.005 \)
- \( (0.01)(0.15) = 0.0015 \)

Now summing these:
\[ P(D) = 0.009 + 0.005 + 0.0015 = 0.0155 \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(E_3 | D) \), the probability that the bulb was produced by machine C given that it is defective:
\[ P(E_3 | D) = \frac{P(D | E_3) P(E_3)}{P(D)} \]
Substituting the known values:
\[ P(E_3 | D) = \frac{(0.01)(0.15)}{0.0155} \]
Calculating the numerator:
\[ (0.01)(0.15) = 0.0015 \]
Thus:
\[ P(E_3 | D) = \frac{0.0015}{0.0155} \]
Calculating this gives:
\[ P(E_3 | D) \approx 0.09677 \]

### Final Answer
The probability that the defective bulb was manufactured by machine C is approximately \( 0.09677 \) or \( 9.68\% \).

---