Question

In: Statistics and Probability

An inspector has the job of checking a screw-making machine at the start of each day....

An inspector has the job of checking a screw-making machine at the start of each day. She finds that the machine needs repairs pairs one day out of ten. When the machine does need repairs, all the screws it makes are defective. Even when the machine is working properly, 5% of the screws it makes are defective; these defective screws are randomly scattered through the day's output. Use a calculator to get approximate answers to the following lowing questions. What is the probability that the machine is in good order if:

a. The first screw the inspector tests is defective?

b. The first two screws are both defective?

c. The first three are all defective?

Solutions

Expert Solution

P(Machine need repair on a any given day) = 1/10

P(Machine need not repait on any given day) = 9/100

Probability that screw is defective on a day when machine is repaired = 1

Probability that screw is defetive on a day when machine is correctly working = 0.05

(a) P(a given screw is defective) = P(Machine need repair on a any given day) * Probability that screw is defective on a day when machine is repaired + P(Machine need not repait on any given day) * Probability that screw is defetive on a day when machine is correctly working

= 1/10 * 1 + 9/10 * 0.05 = 0.145

P(Machine is in good order when one screw is defective) = (9/10 * 0.05)/0.145 = 0.31

(b) P(First two screws are both defective) = P(Machine need repair on a any given day) * Probability that screw is defective on a day when machine is repaired + P(Machine need not repair on any given day) * Probability that screw is defetive on a day when machine is correctly working

= 1/10 * 1 + 9/10 * 0.05 * 0.05 = 0.10225

P(Machine is in good order when two screw is defective) = (9/10 * 0.05 * 0.05)/0.10225 = 0.022

(c) Similarly,

when three are found defective

P(First three screws are both defective) = P(Machine need repair on a any given day) * Probability that screw is defective on a day when machine is repaired + P(Machine need not repair on any given day) * Probability that screw is defetive on a day when machine is correctly working

= 1/10 * 1* 1 * 1 + 9/10 * 0.05 * 0.05 * 0.05 = 0.1001125

P(Machine is in good order when three screw is defective) = (9/10 * 0.05 * 0.05 * 0.05)/0.1001125 = 0.0011

orchestra answered 1 year ago
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