question_answer

  In a factory which manufactures bolts, machines A, B and C manufacture respectively 30%, 50% and 20% of the bolts. Out of their outputs 3%, 4% and 1%, respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.

Answer:

Let\[{{E}_{1}}\]: the event that the item is produced by machine A. \[{{E}_{2}}\]: the event that the item is produced by machine B. and\[{{E}_{3}}\]: the event that the item is produced by machine C. \[\therefore \]      \[P\,({{E}_{1}})=30%=\frac{30}{100},\] \[P\,({{E}_{2}})=50%=\frac{50}{100}\] And      \[P\,({{E}_{3}})=20%=\frac{20}{100}\] Let E : the event that the item chosen is defective \[\therefore \]      \[P\,\left( \frac{E}{{{E}_{1}}} \right)=P\](machine A produced defective items) \[=3%=\frac{3}{100}\] \[P\,\left( \frac{E}{{{E}_{2}}} \right)=P\](machine B produced defective items) \[=4%=\frac{4}{100}\] And \[P\,\left( \frac{E}{{{E}_{3}}} \right)=P\](machine C produced defective items) \[=1%=\frac{1}{100}\] The probability that the selected item was from machine B, given that it is defective, is given by \[P\left( \frac{{{E}_{2}}}{E} \right)\]. By using Baye?s theorem, we get \[P\left( \frac{{{E}_{2}}}{E} \right)=\frac{P\left( \frac{{{E}_{2}}}{E} \right)\,\,P\,({{E}_{2}})}{P\left( \frac{E}{{{E}_{1}}} \right)\,\,P\,({{E}_{1}})+P\,\left( \frac{E}{{{E}_{2}}} \right)+P\,({{E}_{2}})+P\,\left( \frac{E}{{{E}_{3}}} \right)\,\,P\,({{E}_{3}})}\]  \[=\frac{\frac{4}{100}\times \frac{5}{100}}{\frac{3}{100}\times \frac{30}{100}+\frac{4}{100}\times \frac{50}{100}+\frac{1}{100}\times \frac{20}{100}}\] \[=\frac{200}{90+200+20}=\frac{200}{310}=\frac{20}{31}\] \[\therefore \]Probability that the selected was not manufactured by machine \[B=1-P\,\left( \frac{{{E}_{2}}}{E} \right)=1-\frac{20}{31}=\frac{11}{31}\]