Question 997962
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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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \