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

Concept: This is an application of

Bayes' Theorem. Let \[ C=\text{Machine correctly set up} \] \[ I=\text{Machine incorrectly set up} \] \[ A=\text{Two acceptable items are produced} \] Then \[ P(C)=0.8, \qquad P(I)=0.2 \] We need to find \[ P(C|A). \]

Step 1: Find the probability of producing two acceptable items. If the setup is correct, \[ P(A|C)=(0.9)^2=0.81 \] If the setup is incorrect, \[ P(A|I)=(0.4)^2=0.16 \]

Step 2: Apply Bayes' theorem. \[\begin{aligned} P(C|A) &= \frac{P(C)\,P(A|C)} {P(C)\,P(A|C)+P(I)\,P(A|I)} \end{aligned}\] Substituting the values, \[\begin{aligned} P(C|A) &= \frac{0.8\times0.81} {0.8\times0.81+0.2\times0.16} \end{aligned}\] \[\begin{aligned} &= \frac{0.648} {0.648+0.032} \end{aligned}\] \[\begin{aligned} &= \frac{0.648}{0.680} \end{aligned}\] \[\begin{aligned} &= 0.95294 \end{aligned}\] \[\begin{aligned} &\approx 0.95 \end{aligned}\]

Step 3: Write the final answer. \[\begin{aligned} \boxed{0.95} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.