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 \[\frac{2}{5}\].
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