An inspector working for a manufacturing company has a 99% chance of correctly identifying defective items and a 0.5% chance of incorrectly classifying a good item as defective. The company has evidence that its line produces 1.2% of defective items.

Round your answers to five decimal places (e.g. 98.76543).

(a) What is the probability that an item selected for inspection is classified as defective?



(b) If an item selected at random is classified as nondefective, what is the probability that it is indeed good?

Respuesta :

Answer:

a) 1.682%

b.) 99.98779

Step-by-step explanation:

work below

all you really need to know for this question is law of total probability and the conditional probability formula

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Answer:

(a): 1.682 %

(b): 99.98779 %

Step-by-step explanation:

Using the shorthands for the event “item classified as defective” C and “item is defective” D

(a): [tex]P(C) = P(C \cap D) + P(C \cap \overline{D}) = P(D)\cdot P(C | D) + (1-P(D))\cdot P(C | \overline{D}) = 0.012\times 0.99 + 0.988\times 0.005[/tex]

(b): [tex]P(\overline{D}|\overline{C}) = \frac{P(\overline{D} \cap \overline{C})}{P(\overline{C})} = \frac{P(\overline{D})}{P(\overline{C})}\times P(\overline{C} | \overline{D}) = \frac{1 - P(D)}{1 - P(C)}\times (1-P(C | \overline{D})) = \frac{0.988}{0.98318}\times 0.995[/tex]