Question

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipment produced parts are passed through an automatic inspection machine, which is supposed to be able to detect any part that is obviously defective and discard it. However, the inspection machine is not perfect. A part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input. Slightly defective parts are marked as defective and discarded 40% of the time, and obviously defective parts are correctly identified and discarded 98% of the time.
Required:
a. What is the total probability that a part is marked as defective and discarded by the automatic inspection machine?
b. What is the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped?
c. What is the probability that a part is 'bad' (obviously defective) given that it makes it through the inspection machine and gets shipped?

Answer 1

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016