Question

In a manufacturing plant, machine A produces 30% of a certain product, machine B produces 30% of this product, and machine C produces 40% of this product. Six percent of machine A products are defective, 10% of machine B products are defective, and 8% of machine C products are defective. The company inspector has just sampled a product from this plant and has found it to be defective. Determine the revised probabilities that the sampled product was produced by machine A, machine B, or machine C. (Note: A = product produced on Machine A, B = product produces on Machine B, C = product produced on Machine C, D = defective product) (Round your answers to 4 decimal places, e.g. 0.7895.) P(AD) = P(BD) = P(CD) =

Answer 1

P(A|D) = 0.225

P(B|D) = 0.375

P(C|D) = 0.4

Step-by-step explanation:

To determine the revised probabilities that the sampled defective product was produced by machines A, B, or C, we need to use conditional probability.

Let's start by calculating the probability of selecting a defective product from each machine. We are given that 6% of machine A products are defective, 10% of machine B products are defective, and 8% of machine C products are defective.

P(D|A) = 0.06 (probability of selecting a defective product given it was produced by machine A)

P(D|B) = 0.10 (probability of selecting a defective product given it was produced by machine B)

P(D|C) = 0.08 (probability of selecting a defective product given it was produced by machine C)

Now, let's calculate the probability of selecting a product from each machine. We are given that machine A produces 30% of the product, machine B produces 30% of the product, and machine C produces 40% of the product.

P(A) = 0.30 (probability of selecting a product produced by machine A)

P(B) = 0.30 (probability of selecting a product produced by machine B)

P(C) = 0.40 (probability of selecting a product produced by machine C)

To determine the revised probabilities, we need to calculate P(D), which is the probability of selecting a defective product regardless of the machine it was produced on. We can use the law of total probability to find this value.

P(D) = P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C)

= 0.06 * 0.30 + 0.10 * 0.30 + 0.08 * 0.40

= 0.018 + 0.03 + 0.032

= 0.08

Now, we can calculate the revised probabilities using Bayes' theorem. We want to find P(A|D), P(B|D), and P(C|D), which are the probabilities of the product being produced by machine A, B, or C given that it is defective.

P(A|D) = (P(D|A) * P(A)) / P(D)

P(B|D) = (P(D|B) * P(B)) / P(D)

P(C|D) = (P(D|C) * P(C)) / P(D)

Substituting the values we calculated earlier:

P(A|D) = (0.06 * 0.30) / 0.08

= 0.018 / 0.08

= 0.225

P(B|D) = (0.10 * 0.30) / 0.08

= 0.03 / 0.08

= 0.375

P(C|D) = (0.08 * 0.40) / 0.08

= 0.032 / 0.08

= 0.4

Therefore, the revised probabilities are:

P(A|D) = 0.225

P(B|D) = 0.375

P(C|D) = 0.4

Please note that the probabilities are rounded to 4 decimal places.