Use the following problem to answer questions 8 and 9. A television manufacturer has three locations where it is finally assembled. Plants A, B, and C supply 65%, 15%, and 20% respectively of the televisions used by the manufacture. Quality control has determined that 1.5% produced by Plant A are defective, while plants B and C have 2.5% defective products. 8. What is the probability of finding a non-defective television

The probability of finding a non-defective television is 0.9815.

Step-by-step explanation:

We are given that a television manufacturer has three locations where it is finally assembled. Plants A, B, and C supply 65%, 15%, and 20% respectively of the televisions used by the manufacture.

Quality control has determined that 1.5% produced by Plant A are defective, while plants B and C have 2.5% defective products.

Let Probability that Plant A supply televisions = P(A) = 0.65

Probability that Plant B supply televisions = P(B) = 0.15

Probability that Plant C supply televisions = P(C) = 0.20

Also, let D = event that television is defective

Probability that television is defective given that it is supplied by Plant A = P(D/A) = 0.015

Probability that television is defective given that it is supplied by Plant B = P(D/B) = 0.025

Probability that television is defective given that it is supplied by Plant C = P(D/C) = 0.025

Now, to find the probability of finding a non-defective television firstly we will find the probability of a defective television.

Probability of a defective television is given by;

   =  P(A) P(D/A) + P(B) P(D/B) + P(C) P(D/C)  

   =  0.65 0.015 + 0.15 0.025 + 0.20 0.025  

   =  0.00975 + 0.00375 + 0.005

   =  0.0185

Now, probability of finding a non-defective television = 1 - Probability of a defective television

Probability of finding a non-defective television = 1 - 0.0185

                                                                                   = 0.9815

Therefore, the probability of finding a non-defective television is 98.15%.

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step-by-step explanation: