A factory has a total of three manufacturing units M₁, M₂, M₃ producing bulbs independently. Units M

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Question

A factory has a total of three manufacturing units M₁, M₂, M₃ producing bulbs independently. Units M₁, M₂, M₃ produce bulbs in proportions 2:2:1 respectively. 20% of all bulbs produced are defective. 15% of M₁'s bulbs are defective. If a randomly chosen defective bulb was produced by M₂ with probability 2/5, find the probability that a randomly chosen bulb from M₃ is defective.

Accepted Answer

Correct Answer: 0.30Step-by-step Solution:Let P(Eᵢ) = probability bulb is from Mᵢ. Proportions 2:2:1, so:P(E₁) = 2/5, P(E₂) = 2/5, P(E₃) = 1/5.Let P(D|Eᵢ) = probability that a bulb from Mᵢ is defective.Given: P(D|E₁) = 15/100 = 3/20.Total probability of defective:P(D) = P(E₁)P(D|E₁) + P(E₂)P(D|E₂) + P(E₃)P(D|E₃)0.20 = (2/5)(3/20) + (2/5)P(D|E₂) + (1/5)P(D|E₃)0.20 = 6/100 + (2/5)P(D|E₂) + (1/5)P(D|E₃)0.20 − 0.06 = (2/5)P(D|E₂) + (1/5)P(D|E₃)0.14 = (2/5)P(D|E₂) + (1/5)P(D|E₃) ... (1)Using Bayes' theorem for M₂:P(E₂|D) = P(E₂)P(D|E₂) / P(D) = 2/5(2/5)P(D|E₂) / 0.20 = 2/5(2/5)P(D|E₂) = (2/5)(0.20) = 0.08P(D|E₂) = 0.08/(2/5) = 0.08 × 5/2 = 0.20 = 1/5.Wait: (2/5)P(D|E₂)/0.20 = 2/5 → (2/5)P(D|E₂) = 2/5 × 0.20 = 0.08 → P(D|E₂) = 0.08×5/2 = 0.20.Substituting in (1): 0.14 = (2/5)(0.20) + (1/5)P(D|E₃) = 0.08 + (1/5)P(D|E₃)(1/5)P(D|E₃) = 0.06 → P(D|E₃) = 0.30Probability that M₃ bulb is defective = 0.30Key Concepts:Law of Total Probability: P(D) = Σᵢ P(Eᵢ)P(D|Eᵢ). Used when the sample space is partitioned into mutually exclusive events Eᵢ.Bayes' Theorem: P(Eᵢ|D) = P(Eᵢ)P(D|Eᵢ) / P(D). Gives the posterior probability of an event given observed data.Setting up equations from given conditions: Two conditions (total defective = 20%, and P(E₂|D) = 2/5) give two equations in two unknowns P(D|E₂) and P(D|E₃). Always identify the number of unknowns and ensure you have enough equations.Proportions to probabilities: Proportions a:b:c convert to probabilities a/(a+b+c), b/(a+b+c), c/(a+b+c). Here 2:2:1 → 2/5, 2/5, 1/5.

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