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.

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%.

answer: group b.

step-by-step explanation:

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

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