SOLUTION: In a bolt factory, machines A, B and C 25%,35% and 40% of total outputs respectively. Of the total of their outputs, 5%,4% and 2% are defective bolts respectively. A bolt is drawn

Algebra ->  Probability-and-statistics -> SOLUTION: In a bolt factory, machines A, B and C 25%,35% and 40% of total outputs respectively. Of the total of their outputs, 5%,4% and 2% are defective bolts respectively. A bolt is drawn       Log On


   



Question 1193995: In a bolt factory, machines A, B and C 25%,35% and 40% of total outputs respectively. Of the total of their outputs, 5%,4% and 2% are defective bolts respectively. A bolt is drawn at random and is found to be defective. What is the probability that it was manufactured by machines A, B and C?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I'll be picky at first....

Each bolt is produced by a single machine. So the probability that a defective bolts was "manufactured by machines A, B, and C" is zero.

ANSWER: 0

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Now on to answering the question that the problem MEANT to ask....

To avoid working with multiplication of percentages in decimal form, I would choose a convenient number for the total number of bolts produced and work with the resulting whole numbers.

So suppose the total production is 10,000 bolts. Then...

Production:
A: 25% of 10,000 = 2500
B: 35% of 10,000 = 3500
C: 40% of 10,000 = 4000

Faulty bolts:
A: 5% of 2500 = 125
B: 4% of 3500 = 140
C: 2% of 4000 = 80

Total defective bolts: 125+140+80 = 345

P(a defective bolt was produced by machine A) = 125/345
P(a defective bolt was produced by machine B) = 140/345
P(a defective bolt was produced by machine C) = 80/345

Simplify the fractions or convert to decimals as desired/required.