Answer:
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
<em>Also, let D = event that television is defective</em>
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
<u>Now, to find the probability of finding a non-defective television firstly we will find the probability of a defective television.</u>
<em>Probability of a defective television is given by;</em>
= 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, <em><u>probability of finding a non-defective television</u></em> = 1 - Pro<em>bability of a defective television</em>
Probability of finding a non-defective television = 1 - 0.0185
= 0.9815
<em>Therefore, the probability of finding a non-defective television is 98.15%.</em>
