Question 1088140
For the first, 50 of 500 are defective. P(D|FS}=0.1
For the second, 35 of 500 are defective
For the third, 17.5 of 350 are defective
this is 102.5 out of 1350 or 0.0759.
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For the second part, I make a contingency table
========Def=====ND=======Total
First---50------450-------500
Second--35------465-------500
Third---17.5----332.5-----350
Total---102.5---1247.5----1350
If defective, it is 50/102.5 or 0.488 probability it came from first shift.