Before leaving the factory, items are inspected to determine if they are good fo

ID: 3331656 • Letter: B

Question

Before leaving the factory, items are inspected to determine if they are good for use or defective (defective no or yes).

Suppose 10 percent of items actually are defective.

If an item is defective, the inspection process will label it as negative (-) with probability 0.9 (negative yes or no).

If the item is not defective, the inspection process will label it negative (-) with probability 0.05.

Let D denote the event the item is defective, and denote the event the item is labeled as negative.

So we have D for defective and Dc for not defective, and for negative and c (or +) for not negative (positive).

a. What is the probability a randomly chosen item is labeled negative -?

b. An item is labeled negative -. What is the probability that the item is defective

c. Are the events D and independent? Why or why not.

10% of the items are actually defective

so P[D]=0.10

If an item is defective, the inspection process will label it as negative (-) with probability 0.9

so P[-|D]=0.9

If the item is not defective, the inspection process will label it negative (-) with probability 0.05.

so P[-|Dc]=0.05

a) the probability that a randomly chosen item is labeled negative - is

P[-]=P[-|D]*P[D]+P[-|Dc]*P[Dc] [by total probability]

=0.9*0.1+0.05*[1-0.1]=0.135 [answer]

b) P[D|-] can be calculated using Bayes' theorem

P[D|-]=P[-|D]*P[D]/P[-]=0.9*0.1/0.135=0.6667 [answer]

c) no they are not independent. because the conditional distribution of - given D depends on D

P[-|D]=0.9 and P[-|Dc]=0.05 but if they were independent it should be P[-|D]=P[-|Dc]=P[-]

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