The machining center at a production facility has three lathes of differing ages
ID: 3239141 • Letter: T
Question
The machining center at a production facility has three lathes of differing ages and thus precision. The oldest, machine A, produces finished units of product of which 88% are good, 8% are blemished, and 4% unusable. Machine B produces 92% good, 6% blemished, and 2% unusable. The newest machine, machine C, turns out product that is 96% good, 3% blemished, and 1% unusable. If machine A produces 1/4 of the company's output and machine B turns out 1/3 of the output, what fraction of the company's product is good? What percentage is blemished? If an unit of product is selected at random and found to be blemished, what is the probability that it was produced on machine B?
Explanation / Answer
Fraction of the company product is good =
Machine A output * Machine A produces finished goods that are good +
Machine B output * Machine B produces finished goods that are good +
Machine C output * Machine C produces finished goods that are good
1/4 * 0.88 + 1/3 * 0.92 + 5/12 * 0.96
=0.9267
Fraction of the company product is good = 139/150 = 0.9267
Percentage that is blemished is
1/4 * 0.08 + 1/3 * 0.06 + 5/12 * 0.03
=0.0525 = 5.25 %
By Bayes' theorem
p(B/Blemished) = p(Blemished/B) p(B) / [p(Blemished/A) p(A) + p(Blemished/B) p(B) + p(Blemished/C) p(C)]
= 1/3 * 0.06 / (0.0525) (0.0525 as calculated above)
= 0.381
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