SOLUTION: Two machines turn out all the products in a factory, with the first machine producing 57% of the product and the second machine producing the rest. The first machine produces defec

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Question 1203912: Two machines turn out all the products in a factory, with the first machine producing 57% of the product and the second machine producing the rest. The first machine produces defective products 6% of the time and the second machine 2% of the time. What is the probability that a part made in this factory is not defective? (Enter answer as a decimal with at least 4 correct decimal places)
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

A = 1st machine is selected
B = 2nd machine is selected
P(A) = 0.57
P(B) = 1-P(A) = 1-0.57 = 0.43

D = part is defective
~D = part is not defective
P(D given A) = 0.06
P(~D given A) = 1-P(D given A) = 1-0.06 = 0.94
and,
P(D given B) = 0.02
P(~D given B) = 1-P(D given B) = 1-0.02 = 0.98

The four pieces of useful relevant info are
P(A) = 0.57
P(B) = 0.43
P(~D given A) = 0.94
P(~D given B) = 0.98

Then,
P(~D) = P(~D and A) + P(~D and B) ........ Law of total probability
P(~D) = P(~D given A)*P(A) + P(~D given B)*P(B)
P(~D) = 0.94*0.57 + 0.98*0.43
P(~D) = 0.9572 which is the final answer.
There's a 95.72% chance the part is not defective.

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An alternative approach:

Let's say that 100,000 parts were made.
Machine A makes 57,000 of those parts (because 57% of 100,000 = 57,000). The remaining 100,000-57,000 = 43,000 parts are from machine B.
Note how 43% of 100,000 is 43,000.

Machine A messes up 6% of the time.
6% of 57,000 = 0.06*57000 = 3420
There are 3420 defective parts made from machine A. The remaining 57,000-3,420 = 53,580 parts are not defective.
Or we could say: 94% of 57,000 = 0.94*57000 = 53,580.

Machine B messes up 2% of the time.
2% of 43,000 = 0.02*43000 = 860
There are 860 defective parts made from machine B. The remaining 43,000-860 = 42,140 parts are not defective.
Or we could say: 98% of 43,000 = 0.98*43000 = 42,140.

In total, there are 53580+42140 = 95,720 non-defective parts out of 100,000 parts total.

Here's a table to summarize everything stated so far in this section.
DefectiveNot DefectiveTotal
Machine A3,42053,58057,000
Machine B86042,14043,000
Total4,28095,720100,000

The probability of randomly selecting a non-defective part is (95,720)/(100,000) = 0.9572

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Answer: 0.9572
Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.
Two machines turn out all the products in a factory, with the first machine producing 57% of the product
and the second machine producing the rest.
The first machine produces defective products 6% of the time and the second machine 2% of the time.
What is the probability that a part made in this factory is not defective?
(Enter answer as a decimal with at least 4 correct decimal places)
~~~~~~~~~~~~~~~~~~~~~~~~~ 

If to think one - two minutes, the solution, the formula and answer can be written in one line

    P = 0.57*(1-0.06) + (1-0.57)*(1-0.02) = 0.9572.    ANSWER

Solved.

The formula is SELF-EXPLANATORY.

Look in it attentively, and it will tell you a whole story.



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