question_answer

        In an electric bulb factory, machines A, B and C manufacture 60%, 30% and 10% electric bulbs, respectively. 1%, 2% and 3% of the electric bulbs produced respectively by A, B and C are found to be defective. An electric bulb is picked up at random from the total production and found to be defective. Find the probability that defective electric bulb was produced by machine A.

Answer:

Let \[{{E}_{1,}}\]\[{{E}_{2}}\]and \[{{E}_{3}}\] be the events to manufacture electric bulbs by A, B and C, respectively. Then, \[P\,({{E}_{1}})=60%=\frac{60}{100}=\frac{6}{10}\] \[P\,({{E}_{2}})=30%=\frac{30}{100}=\frac{3}{10}\] And \[P\,({{E}_{3}})=10%=\frac{10}{100}=\frac{1}{10}\] Let \[E\] be the event of selecting a defective electric bulb, then \[P\,(E/{{E}_{1}})=P\](selecting a defective electric bulb from machine A) \[=1%=\frac{1}{100}\] \[P\,(E/{{E}_{2}})=P\](selecting a defective electric bulb from machine B) \[=2%=\frac{2}{100}\] and \[P\,(E/{{E}_{3}})=P\](selecting a defective electric bulb from machine C) \[=3%=\frac{3}{100}\] \[\therefore \]Probability that the selected defective electric bulb was produced by machine A, [using Baye?s theorem] \[=\frac{\frac{6}{10}\times \frac{1}{100}}{\left( \frac{6}{10}\times \frac{1}{100} \right)+\left( \frac{3}{10}\times \frac{2}{100} \right)+\left( \frac{1}{10}\times \frac{3}{100} \right)}\] \[=\frac{6}{6+6+3}=\frac{6}{15}=\frac{2}{5}\] Hence, the probability that the selected defective electric bulb was produced by machine A, is \[\frac{2}{5}\].