Three factories produce light bulbs to supply the market. Factory A produces 40%, 20% of the tools are produced in factory B and 40% in fact

Question

Three factories produce light bulbs to supply the market. Factory A produces 40%, 20% of the tools are produced in factory B and 40% in factory C. 5% of the bulbs produced in factory A, 3% of the bulbs produced in factory B and 4% of the bulbs produced in factory C are defective.A bulb is selected at random in the market and found to be defective. What is the probability that this bulb was produced by factory A?

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MichaelMet 3 years 2021-08-22T13:36:53+00:00 1 Answers 95 views 0

Answers ( )

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    2021-08-22T13:38:11+00:00

    Answer:

    0.4762 = 47.62% probability that this bulb was produced by factory A

    Step-by-step explanation:

    Conditional Probability

    We use the conditional probability formula to solve this question. It is

    P(B|A) = \frac{P(A \cap B)}{P(A)}

    In which

    P(B|A) is the probability of event B happening, given that A happened.

    P(A \cap B) is the probability of both A and B happening.

    P(A) is the probability of A happening.

    In this question:

    Event A: Bulb is defective.

    Event B: Produced by factory A.

    Probability of a bulb being defective.

    5% of 40%(factory A)

    3% of 20%(factory B)

    4% of 40%(factory C). So

    P(A) = 0.05*0.4 + 0.03*0.2 + 0.04*0.4 = 0.042

    Defective and from factory A:

    5% of 40%. So

    P(A \cap B) = 0.05*0.4 = 0.02

    What is the probability that this bulb was produced by factory A?

    P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.042} = 0.4762

    0.4762 = 47.62% probability that this bulb was produced by factory A

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