Question 1203912
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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) = <font color=red size=4>0.9572</font> 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.
<table border = "1" cellpadding = "5"><tr><td></td><td>Defective</td><td>Not Defective</td><td>Total</td></tr><tr><td>Machine A</td><td>3,420</td><td>53,580</td><td>57,000</td></tr><tr><td>Machine B</td><td>860</td><td>42,140</td><td>43,000</td></tr><tr><td>Total</td><td>4,280</td><td>95,720</td><td>100,000</td></tr></table>
The probability of randomly selecting a non-defective part is (95,720)/(100,000) = <font color=red size=4>0.9572</font>


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Answer: <font color=red size=4>0.9572</font>
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