Question

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

Answer 1

Answer:

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