The factory produces batches of 100 components, with 60 Type A and 40 Type B. The labeling machine mislabels components: Type A is mislabeled as Type B with probability 0.1, and Type B is mislabeled as Type A with probability 0.2. However, the test procedure is independent of the labels and directly measures the true type of the component with 0.95 accuracy. The goal is to find the probability that a component is actually Type A given that the test says it is Type A, denoted as (P(\text{Actually A} \mid \text{Test says A})).

• Let (A) be the event that the component is actually Type A.
• Let (B) be the event that the component is actually Type B.
• Let (T_A) be the event that the test says the component is Type A.
• Let (T_B) be the event that the test says the component is Type B.

Step 2: Prior Probabilities

From the batch composition:

• (P(A) = \frac{60}{100} = 0.6)
• (P(B) = \frac{40}{100} = 0.4)

Step 3: Test Probabilities

The test correctly identifies the true type with probability 0.95:

• (P(T_A \mid A) = 0.95) (test says A given actually A)
• (P(T_B \mid A) = 0.05) (test says B given actually A)
• (P(T_A \mid B) = 0.05) (test says A given actually B)
• (P(T_B \mid B) = 0.95) (test says B given actually B)

Step 4: Apply Bayes' Theorem

Bayes' theorem states: [ P(A \mid T_A) = \frac{P(T_A \mid A) \cdot P(A)}{P(T_A)} ] where (P(T_A)) is the total probability that the test says A.

Step 5: Calculate (P(T_A))

[ P(T_A) = P(T_A \mid A) \cdot P(A) + P(T_A \mid B) \cdot P(B) ] [ P(T_A) = (0.95 \cdot 0.6) + (0.05 \cdot 0.4) = 0.57 + 0.02 = 0.59 ]

Step 6: Calculate (P(A \mid T_A))

[ P(A \mid T_A) = \frac{0.95 \cdot 0.6}{0.59} = \frac{0.57}{0.59} = \frac{57}{59} ]

Step 7: Simplify the Fraction

[ \frac{57}{59} \approx 0.9661 ] The fraction (\frac{57}{59}) is already in simplest form.

Why the Labeling Machine is Irrelevant

The labeling machine mislabels components, but the test procedure is independent of the labels. The test directly measures the true type with 0.95 accuracy, so the mislabeling does not affect the test result. Thus, the labeling machine's probabilities are not used in this calculation.

Final Answer

The probability that the component is actually Type A given that the test says it is Type A is (\frac{57}{59}).