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19 June, 02:13

A chip company has two manufacturing plants. Plant A produces 40% of the chips and Plant B produces 60% of the chips produced by the company. The company knows that 2% of the chips produced by plant A are defective and 1% of the chips produced by plant B are defective. If a randomly chosen chip produced by the company is defective, what is the likelihood that the chip came from plant A

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  1. 19 June, 02:37
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    P (A/D) = 0.5714

    Step-by-step explanation:

    Let's call A the event that a chip is produced by Plant A, B the event that a chip is produced by Plant B and D the event that the chip is defective

    So, the likelihood or probability P (A/D) that a chip came from plant A given that the chip is defective is calculated as:

    P (A/D) = P (A∩D) / P (D)

    Where P (D) = P (A∩D) + P (B∩D)

    Then, the probability P (A∩D) that a chip is produced by plant A and it is defective is calculated as:

    P (A∩D) = 0.4*0.02 = 0.008

    Because, Plant A produces 40% of the chips and 2% of the chips produced by plant A are defective.

    At the same way, the probability P (B∩D) that a chip is produced by plant B and it is defective is calculated as:

    P (B∩D) = 0.6*0.01 = 0.006

    So, P (D) and P (A/D) are equal to:

    P (D) = 0.008 + 0.006 = 0.014

    P (A/D) = 0.008/0.014 = 0.5714

    it means that if a randomly chosen chip produced by the company is defective, the likelihood that the chip came from plant A is 0.5714
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