SOLUTION: In a certain assembly plant, three machines, A, B, and C, make 34%, 20%, and 46%, respectively, of the products. It is known from past experience that 3%, 3%, and 2% of the product

Question 997962: In a certain assembly plant, three machines, A, B, and C, make 34%, 20%, and 46%, respectively, of the products. It is known from past experience that 3%, 3%, and 2% of the products made by each machine, respectively are defective.
a) Draw a tree diagram of the scenario.
b) If a product is chosen at random, what is the probability that it is defective?
c) Given a randomly selected product was found to be defective, what is the probability it was made by machine C?
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I know it said to not post multiple questions. But, I am having such a hard time with this class. I am in ST 314. In class I get where she is headed, but then at home it just doesn't make any sense. I feel like there is a million different formulas and I don't know which one to use.
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A) I know what the diagram looks like. I looked at her example on line but I don't see how that relates to this problem at all.
B) I feel like you have to add together somehow the 34%, the 20%, and the 46% in order to get a correct probability.
C) Then I assume this is similar to part B.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Part B. Pct made on A times pct defective on A PLUS Pct made on B times pct defective on B PLUS Pct made on C times pct defective on C.

Part C. Bayes Theorem. The probability of machine C given part Defective is the (Probability part made on C TIMES Probability part defective given it was made on C) DIVIDED by the Probability the part is defective at all.

So, (.46 times .02)/(The answer to part B)

John

My calculator said it, I believe it, that settles it

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