In a factory there are three separate production lines which all produce the same type of CD-player. Line A produces 30% of the CD-players, line B produces 25% and line C 45%. In a quality inspection 2% of line A's players are faulty, in line B 3% are faulty and in line C 5%.
One CD-player is picked at random. What is the probability that it is:
b) It has been produced in line B when it is know that it is faulty?
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Stacey Warren - Expert brainly.com
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I believe that Venn diagram and Bayes' theorem are of use in this task.
Using the law of total probability:
P[Faulty] = P[Faulty|Factory A]P[Line A] + P[Faulty|Line B]P[Line B] + P[Faulty|Line C]P[Line C]
\[.02 \times .3 + .03 \times .25 + .05 \times .45\]
.036 or 3.6% chance
I believe that b) can now be solved using Bayes' theorem. Let F = faulty and B = line B. Now Bayes' theorem is: \[P(B|F)=(P(B)P(F|B))/P(F)\]
Plug in the values we allready have we get: