Question

Companies A, B, and C produce 40%,20%, and 40%, respectively, of the major appliances sold in a certain area. In that area, 3% of the company A appliances, 121​% of the company B appliances, and 1% of the company C appliances need service within the first year. Suppose a defective appliance is chosen at random; find the probability that it was manufact by The probability that it came from company B is (Type an integer or decimal rounded to four decimal places as needed.)

Answer 1

Answer: To find the probability that a defective appliance chosen at random came from company B, we need to calculate the conditional probability.

Let's denote the events as follows:

A: Appliance is from company A

B: Appliance is from company B

C: Appliance is from company C

D: Appliance needs service within the first year

We are given the following probabilities:

P(A) = 0.40 (company A's market share)

P(B) = 0.20 (company B's market share)

P(C) = 0.40 (company C's market share)

P(D|A) = 0.03 (probability that a company A appliance needs service)

P(D|B) = 1.21 (probability that a company B appliance needs service)

P(D|C) = 0.01 (probability that a company C appliance needs service)

We want to find P(B|D), the probability that a defective appliance came from company B.

By applying Bayes' theorem, we have:

P(B|D) = (P(D|B) * P(B)) / [P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C)]

Substituting the given values, we get:

P(B|D) = (1.21 * 0.20) / [(0.03 * 0.40) + (1.21 * 0.20) + (0.01 * 0.40)]

Calculating this expression, we find:

P(B|D) ≈ 0.2449

Therefore, the probability that a defective appliance chosen at random was manufactured by company B is approximately 0.2449 (rounded to four decimal places).

Explanation: