A manufacturer has three machine operators \mathrm{A}, \mathrm{B} and \mathrm{C}. The first operator \mathrm{A} produces 1 \% defective items, whereas the other two operators B and C produce 5 \% and 7 \% defective items respectively. A is on the job for 50 \% of the time, B is on the job for 30 \% of the time and \mathrm{C} is on the job for 20 \% of the time. A defective item is produced, what is the probability that it was produced by A?
A manufacturer has three machine operators \mathrm{A}, \mathrm{B} and \mathrm{C}. The first operator \mathrm{A} produces 1 \% defective items, whereas the other two operators B and C produce 5 \% and 7 \% defective items respectively. A is on the job for 50 \% of the time, B is on the job for 30 \% of the time and \mathrm{C} is on the job for 20 \% of the time. A defective item is produced, what is the probability that it was produced by A?

Solution:

Let \mathrm{E}_{1} be the event of machine time consumption. be the event of machine time consumption \mathrm{A}, \mathrm{E}_{2} Let B and E_{3} represent the event of machine C using time. Let X be the occurrence of defective items being produced.

Then P\left(E_{1}\right)=50 \%=50 / 100=1 / 2

P\left(E_{2}\right)=30 \%=30 / 100=3 / 10

P\left(E_{3}\right)=20 \%=20 / 100=1 / 5

Because a headed coin has a head on both sides, it will display one.

Also P\left(X \mid E_{1}\right)=P( defective item produced by A)=1 \%=1 / 100

And P\left(X \mid E_{2}\right)=P( defective item produced by B)=5 \%=5 / 100

And P\left(X \mid E_{3}\right)=P( defective item produced by C)=7 \%=7 / 100

Now, given that a defective item is generated, the probability of an item manufactured by machine A is P\left(E_{1} \mid A\right).

We have used Bayes’ theorem to arrive at our conclusion.

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{X}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{3}\right)}

We may now retrieve the result by swapping the values.

=\frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{2} \cdot \frac{1}{100}+\frac{3}{10} \cdot \frac{5}{100}+\frac{1}{5} \cdot \frac{7}{100}}

=\frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{100}\left(\frac{1}{2}+\frac{3}{2}+\frac{7}{5}\right)}

=\frac{\frac{1}{2}}{\frac{17}{5}}=\frac{5}{34}

P\left(E_{1} \mid A\right)=5 / 34