Question

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 0.7% of nonconforming 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?

Answer 1

Answer:

(a) The probability that an item selected for inspection is classified as defective is 0.01189.

(b) The probability that an item selected at random is classified as non-defective when in fact it is good is 0.99992.

Step-by-step explanation:

Let's denote the events as follows:

G = an item is good

B = an item is bad

D = an item is classified as defective.

Given:

$P (D|B) = 0.99\\P(D|G)=0.005\\P(B)=0.007$

The probability of producing good items is:

$P(G)=1-P(B)=1-0.007=0.993$

(a)

The law of total probability states that:

$P(A)=P(A|B)P(B)+P(A|C)P(C)$

Using the law of total probability determine the probability that an item selected for inspection is classified as defective as follows:

$P(D)=P(D|G)P(G)+P(D|B)P(B)\\=(0.005\times0.993)+(0.99\times0.007)\\=0.01189$

Thus, the probability that an item selected for inspection is classified as defective is 0.01189.

(b)

Compute the probability that an item selected at random is classified as non-defective when in fact it is good as follows:

$P(G|D^{c})=\frac{P(D^{c}|G)P(G)}{P(D^{c})} \\=\frac{[1-P(D|G)]P(G)}{1-P(D)} \\=\frac{[1-0.005]\times0.993}{1-0.01189} \\=0.99992$

Thus, the probability that an item selected at random is classified as non-defective when in fact it is good is 0.99992.