Question

A factory has a total of three manufacturing units, $ M_1, M_2, M_3 $, which produce bulbs independently of each other. The units $ M_1, M_2, M_3 $ produce bulbs in the proportions $ 2 : 2 : 1 $, respectively. It is known that 20% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $ M_1 $, 15% are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $ M_2 $ is $ \frac{2}{5} $. If a bulb is chosen randomly from the bulbs produced by $ M_3 $, then the probability that it is defective is ________.

Hint:

Use Bayes’ Theorem when conditional and reverse conditional probabilities are involved. Total probability theorem helps combine probabilities across multiple cases.

Answer 1

Step 1: Let the total production be in the ratio \( M_1 : M_2 : M_3 = 2 : 2 : 1 \) Let total production = 5 units
Then: - \( P(M_1) = \frac{2}{5} \) - \( P(M_2) = \frac{2}{5} \) - \( P(M_3) = \frac{1}{5} \) Let: - \( D \): event that a bulb is defective - \( P(D|M_1) = 0.15 \) - \( P(D|M_2) = x \) (unknown) - \( P(D|M_3) = y \) (we need to find this) We are told: - \( P(D) = 0.20 \) - \( P(M_2|D) = \frac{2}{5} \)
Step 2: Use total probability theorem: \[ P(D) = P(M_1) \cdot P(D|M_1) + P(M_2) \cdot P(D|M_2) + P(M_3) \cdot P(D|M_3) \Rightarrow 0.20 = \frac{2}{5} \cdot 0.15 + \frac{2}{5} \cdot x + \frac{1}{5} \cdot y \] \[ \Rightarrow 0.20 = 0.06 + \frac{2x}{5} + \frac{y}{5} \Rightarrow 0.14 = \frac{2x + y}{5} \Rightarrow 2x + y = 0.70 \quad \text{(i)} \]
Step 3: Use Bayes' Theorem: \[ P(M_2|D) = \frac{P(M_2) \cdot P(D|M_2)}{P(D)} = \frac{\frac{2}{5} \cdot x}{0.20} = \frac{2x}{1} \Rightarrow \frac{2x}{1} = \frac{2}{5} \Rightarrow x = \frac{1}{5} \] Now plug into equation (i): \[ 2 \cdot \frac{1}{5} + y = 0.70 \Rightarrow \frac{2}{5} + y = 0.70 \Rightarrow y = 0.70 - 0.4 = \boxed{0.3} \] Wait, we made an arithmetic mistake here. Actually: \[ 2x = \frac{2}{5} \Rightarrow x = \frac{1}{5} \] Then from (i): \[ 2 \cdot \frac{1}{5} + y = 0.70 \Rightarrow \frac{2}{5} + y = 0.70 \Rightarrow y = 0.70 - 0.4 = \boxed{0.3} \] So \( P(D|M_3) = \boxed{0.3} \) Wait — the problem is asking for \( P(D|M_3) \) if \( P(M_2|D) = \frac{2}{5} \) Let’s recompute carefully. Go back to: \[ 0.20 = \frac{2}{5} \cdot 0.15 + \frac{2}{5} \cdot x + \frac{1}{5} \cdot y \Rightarrow 0.20 = 0.06 + \frac{2x}{5} + \frac{y}{5} \Rightarrow 0.14 = \frac{2x + y}{5} \Rightarrow 2x + y = 0.70 \quad \text{(1)} \] Bayes’ theorem: \[ P(M_2|D) = \frac{\frac{2}{5} \cdot x}{0.20} = \frac{2x}{1} = \frac{2}{5} \Rightarrow 2x = \frac{2}{5} \Rightarrow x = \frac{1}{5} \] Now use in (1): \[ 2 \cdot \frac{1}{5} + y = 0.70 \Rightarrow y = 0.70 - 0.4 = \boxed{0.3} \] So the required answer is: \[ \boxed{0.3} \]