Three machines A, B, C produce respectively 50%, 30%, and 20% of the total no. of items of a factory. The % of defective outputs of these machines is 3%, 4%, 5% respectively. If an item is selected at random, (i) what is the probability that it is defective?

Answer

To find the probability that an item selected at random is defective, we will follow these steps:

### Step 1: Define the events and their probabilities
Let:
- \( E_1 \): Event that an item is produced by machine A
- \( E_2 \): Event that an item is produced by machine B
- \( E_3 \): Event that an item is produced by machine C

The probabilities of these events based on their production percentages are:
- \( P(E_1) = 0.50 \) (50% of items produced by A)
- \( P(E_2) = 0.30 \) (30% of items produced by B)
- \( P(E_3) = 0.20 \) (20% of items produced by C)

### Step 2: Define the conditional probabilities of defective items
Let \( A \) be the event that an item is defective. The conditional probabilities of an item being defective given the machine that produced it are:
- \( P(A | E_1) = 0.03 \) (3% defective from machine A)
- \( P(A | E_2) = 0.04 \) (4% defective from machine B)
- \( P(A | E_3) = 0.05 \) (5% defective from machine C)

### Step 3: Use the law of total probability
The total probability of an item being defective can be calculated using the formula:
\[ P(A) = P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2) + P(A | E_3) \cdot P(E_3) \]

### Step 4: Substitute the values into the formula
Substituting the known values:
\[ P(A) = (0.03 \cdot 0.50) + (0.04 \cdot 0.30) + (0.05 \cdot 0.20) \]

### Step 5: Calculate each term
Calculating each term:
- \( 0.03 \cdot 0.50 = 0.015 \)
- \( 0.04 \cdot 0.30 = 0.012 \)
- \( 0.05 \cdot 0.20 = 0.01 \)

### Step 6: Sum the terms
Now, add these results together:
\[ P(A) = 0.015 + 0.012 + 0.01 = 0.037 \]

### Final Result
Thus, the probability that an item selected at random is defective is:
\[ P(A) = 0.037 \quad \text{or} \quad 3.7\% \]

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