If a machine is correctly set up, it produces 90% Acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Past experiences show that 80% of the setups are correctly done. if after a certain setup, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.

Hint: We need the probability that the machine is correctly set up. So, use the concept of conditional probability to find the probabilities of producing 2 acceptable items when it is given a correctly set up machine and also the probability of producing 2 items when it is given that machine is in incorrect setup. If three events A,B,C are such that $A\cup B=1$ and $A\cap B=0$ then we can write, the below formula:
\[P\left( A/C \right)=\dfrac{P\left( A \right)\times P\left( C/A \right)}{P\left( A \right)\times P\left( C/A \right)+P\left( B \right)\times P\left( C/B \right)}\]

Complete step-by-step answer:
Let us assume following three conditions before solving the equation:
${{E}_{1}}$: Event that machine is in the correct setup.
${{E}_{2}}$: Event that the machine is in incorrect setup.
$A$: Event that the machine produces two acceptable items.

By basic knowledge of probability, if three events A,B,C all such that $A\cup B=1$ and $A\cap B=0$ then we write formula:
$P\left( A/C \right)=\dfrac{P\left( A \right)\times P\left( C/A \right)}{P\left( A \right)\times P\left( C/A \right)+P\left( B \right)\times P\left( C/B \right)}$
Here we have,
$\begin{align}
  & A={{E}_{1}},\,\,B={{E}_{2}},\,\,C=A \\
 & \\
\end{align}$
By substituting these in above formula, we convert it into:
$P\left( {{E}_{1}}/A \right)=\dfrac{P\left( {{E}_{1}} \right)\times P\left( A/{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)\times P\left( A/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)\times P\left( A/{{E}_{2}} \right)}.......(i)$
For solving above equation, we need value of $P\left( {{E}_{1}} \right),P\left( {{E}_{2}} \right),P\left( A/{{E}_{1}} \right),P\left( A/{{E}_{2}} \right)$, for these values we have following cases:
Case 1: Solving for the value of $P\left( {{E}_{1}} \right)$: Probability of correct setup
$P\left( {{E}_{1}} \right)=80\%=\dfrac{80}{100}$
Case 2: Solving for the value of $P\left( {{E}_{2}} \right)$: Probability of machine with incorrect setup.
$P\left( {{E}_{1}} \right)=\left( 100-80 \right)\%=20\%$
By simplifying above we can say that $P\left( {{E}_{2}} \right)=0.2$
Case 3: Solving for $P\left( A/{{E}_{1}} \right)$: Probability of 2 acceptable items if it is given it has the correct setup. Given that it is 90% for producing acceptable items with correct setup.
It is given for 1 item but we have 2 items. So, we get:
$P\left( A/{{E}_{1}} \right)=90\% \times 90\%=0.9\times 0.9$
By simplifying above we can say that
$P\left( A/{{E}_{1}} \right)=0.81$
Case 4: solving for $P\left( A/{{E}_{2}} \right)$: Probability of 2 acceptable items with incorrect setup.
Given that it is 40% for producing acceptable items with incorrect setup.
It is given for 1 item but we have 2 items. So, we get:
$P\left( A/{{E}_{2}} \right)=40\% \times 40\%=0.4\times 0.4$
By simplifying the above equation, we can say value of the $P\left( A/{{E}_{2}} \right)$ =0.16
 By substituting these values into equation (i), we get:
$P\left( {{E}_{1}}/A \right)=\dfrac{0.8\times 0.81}{0.8\times 0.81+0.2\times 0.16}=\dfrac{0.648}{0.680}$
By solving above equation, we get the value as follows:
$P\left( {{E}_{1}}/A \right)=0.95$
Therefore, the probability of the machine being correctly set up if it produces 2 acceptable items is 0.95.

Note: Be careful while applying the formula of $P\left( {{E}_{1}}/A \right)$ as if you write representation wrongly you may lead to different answers. Don’t forget to multiply the terms 2 times in case 3, 4 as the given value is for 1 item. If you don’t take this step you may lead to the wrong answer.