Question

If a machine is correctly set up, it produces \(90\%\) acceptable items. If it is incorrectly set up, it produces \(40\%\) acceptable items. Past experience shows that \(80\%\) of the setups are correctly done. If after a certain setup, the machine produces \(2\) acceptable items, then the probability that the machine is correctly set up is

• A. \(0.85\)
• B. \(0.95\)
• C. \(0.75\)
• D. \(0.65\)

Hint:

Bayes' Theorem: \[ P(A|B) = \frac{P(A)\,P(B|A)} {P(A)\,P(B|A)+P(A^c)\,P(B|A^c)} \] Always identify:

• Prior probabilities

• Conditional probabilities

• Required posterior probability
before substituting values.

Answer 1

The Correct Option is B

Step 1: Name the events.
Let $C$ be a correct setup and $I$ an incorrect setup, with $P(C)=0.8$ and $P(I)=0.2$. Let $A$ be the event that two items in a row are acceptable.

Step 2: Acceptance chance for each setup.
A correct machine gives an acceptable item with probability $0.9$, so two in a row is $0.9\times0.9=0.81$. An incorrect machine gives $0.4$, so two in a row is $0.4\times0.4=0.16$.

Step 3: Weight each by its setup chance.
Correct branch: $P(C)\,P(A\mid C)=0.8\times0.81=0.648$. Incorrect branch: $P(I)\,P(A\mid I)=0.2\times0.16=0.032$.

Step 4: Total chance of two acceptable items.
$P(A)=0.648+0.032=0.680$.

Step 5: Apply the reverse-probability idea.
We want the chance the setup was correct given the good output, which is the correct branch divided by the total: $P(C\mid A)=\dfrac{0.648}{0.680}$.

Step 6: Simplify.
$\dfrac{0.648}{0.680}=0.9529\ldots\approx 0.95$. \[ \boxed{0.95} \]