Question

In a certain assembly plant, three machines B1, B2, B3 make 35%, 45% and 20% respectively, of the products. It is known that 2%, 3% and 2% of the products made by each machine, respectively, are defective. Now suppose that a finished product is randomly selected. Obtain the probability that it is defective?

Answer 1

The probability of the defective product is 0.305

Given:

Products produced by plant B1 are 35%

Products produced by plant B2 are 45%

Products produced by plant B3 are 20%

Defective products of plant B1 are 2%

Defective products of plant B2 is 3%

Defective products of plant B3 is 2%

To Find:

The probability of the defective product

Solution:

We use the Theorem of Total probability which is expressed as

If the events B1...., BK  form a partition of the sample space such that P($B_{i}$) ≠ = 0, i = 1,...., K, then for any event A in the sample space S :

P(A) = P($B_{i}$ n A) where i = 1,....,K

Products produced by plant B1 are 35% i.e. P(B1) = 0.35

Products produced by plant B2 are 45% i.e. P(B2) = 0.45

Products produced by plant B3 are 20%  i.e. P(B3) = 0.5

Defective products of plant B1 are 2% i.e. P(D/B1) = 0.2

Defective products of plant B2 is 3% i.e. P(D/B2) = 0.3

Defective products of plant B3 is 2% i.e. P(D/B3) = 0.2

Using Bayes Theorem, we have  

P(D) = {P(D/B1) × P(B1)} + {P(D/B2) × P(B2)} + {P(D/B3) × P(B1)}

       = (0.2 × 0.35) + (0.3 × 0.45) + (0.2 × 0.5)

       = 0.07 + 0.135 + 0.1

       = 0.305

The probability of the defective product is 0.305

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