## anonymous 5 years ago 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: a) Faulty b) It has been produced in line B when it is know that it is faulty? Thank you for your help!

1. anonymous

I believe that Venn diagram and Bayes' theorem are of use in this task.

2. anonymous

a) Faulty 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

3. anonymous

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: $P(B|F)=(0.25\times0.03)/0.036$ $P(B|F)\approx0.21$