Question

In a certain assembly plant, three machines, B1, B2, B3, make 30%, 45%, and 25%, respectively, of the products. It is known from past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective.
A. Suppose that a finished product is randomly selected. What is the probability that it is defective?
B. If a product were chosen randomly and found to be defective, what is the probability that it was made by machine B3.

Answer 1

Answer:

a. The probability that the finished product selected is defective is 2.45%.

b. The probability that product chosen randomly was defective and made by machine B3 is 20.41%.

Step-by-step explanation:

Let A represents the defective product.

We also have the following from the question:

P(B1) = Probability or percentage of the made by machine B1 = 30%, or 0.30

P(B2) = Probability or percentage of the made by machine B2 = 45%, or 0,45

P(B3) = Probability or percentage of the made by machine B3 = 25%, or 0.25

P(A/B1) = Probability or percentage of product B1 that is defective = 2%, or 0.02

P(A/B2) = Probability or percentage of product B2 that is defective = 3%, or 0.03

P(A/B3) = Probability or percentage of product B3 that is defective = 2%, or 0.02

We can therefore proceed as follows:

A. Suppose that a finished product is randomly selected. What is the probability that it is defective?

To determine this, the rules of elimination is applied and we have:

P(A) = (P(B1) * P(A/B1)) + (P(B2) * P(A/B2)) + (P(B3) * P(A/B3)) ………… (1)

Where;

P(A) = Probability that the selected product is defective = ?

Substitutes the values defined above into equation (1), we have:

P(A) = (0.30 * 0.02) + (0.45 * 0.03) + (0.25 * 0.02)

P(A) = 0.006 + 0.0135 + 0.005

P(A) = 0.0245, or 2.45%

Therefore, the probability that the finished product selected is defective is 2.45%.

B. If a product were chosen randomly and found to be defective, what is the probability that it was made by machine B3.

To calculate this, the Bayes’ rule is employed as follows:

P(B3/A) = (P(B3) * P(A/B3)) / [(P(B1) * P(A/B1)) + (P(B2) * P(A/B2)) + (P(B3) * P(A/B3))] = (P(B3) * P(A/B3)) / P(A)  ………….... (2)

Where;

P(B3/A) = The probability that product chosen randomly was defective and made by machine B3 = ?

Also, from the values already defined and obtained in part A, we have:

P(B3) = 0.25

P(A/B3) = 0.002

P(A) = 0.0245

Substituting the values into equation (2), we have:

P(B3/A) = (0.25 * 0.02) / 0.0245

P(B3/A) = 0.005 / 0.0245

P(B3/A) = 0.2041, or 20.41%

Therefore, the probability that product chosen randomly was defective and made by machine B3 is 20.41%.