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Question
If a machine is correctly set up it produces 90% acceptable items. If it is incorrectly set up it produces only 40% acceptable item. Past experience shows that 80% of the setups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.
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Solution
Let A be the event that the machine produces two acceptable items.
Also, let E1 represent the event that the machine is correctly set up and E2 represent the event that the machine is incorrectly set up
\[\therefore P\left( E_1 \right) = 0 . 8 \]
\[ P\left( E_2 \right) = 0 . 2\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 0 . 9 \times 0 . 9 = 0 . 81\]
\[P\left( A/ E_2 \right) = 0 . 40 \times 0 . 40 = 0 . 16\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{0 . 8 \times 0 . 81}{0 . 8 \times 0 . 81 + 0 . 2 \times 0 . 16}\]
\[ = \frac{81}{85}\]
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