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](https://tex.z-dn.net/?f=P%5Cleft%28%5Cfrac%7BE%7D%7BA%7D%5Cright%29%3D0.02)
Now, from the conditional probability,
![P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}](https://tex.z-dn.net/?f=P%5Cleft%28%5Cfrac%7BE%7D%7BA%7D%5Cright%29%3D%5Cfrac%7BP%28E%5Ccap%20A%29%7D%7BP%28A%29%7D)
![\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%5Ccap%20A%29%3DP%5Cleft%28%5Cfrac%7BE%7D%7BA%7D%5Cright%29%5Ctimes%20P%28A%29)
![\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%5Ccap%20A%29%3D0.02%5Ctimes%200.90%3D0.018%5Ccdots%28i%29)
This is the probability of disregarding the defect-free parts by inspection machine.
Similarly,
![P\left(\frac{E}{A}\right)=0.40](https://tex.z-dn.net/?f=P%5Cleft%28%5Cfrac%7BE%7D%7BA%7D%5Cright%29%3D0.40)
and ![\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%5Ccap%20B%29%3D0.40%5Ctimes%200.03%3D0.012%5Ccdots%28ii%29)
This is the probability of disregarding the partially defective parts by inspection machine.
![P\left(\frac{E}{A}\right)=0.98](https://tex.z-dn.net/?f=P%5Cleft%28%5Cfrac%7BE%7D%7BA%7D%5Cright%29%3D0.98)
and ![\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%5Ccap%20C%29%3D0.98%5Ctimes%200.07%3D0.0686%5Ccdots%28iii%29)
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)](https://tex.z-dn.net/?f=%3DP%28E%5Ccap%20C%29)
[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)](https://tex.z-dn.net/?f=P%28E%29%3DP%28E%5Ccap%20A%29%2BP%28E%5Ccap%20B%29%2BP%28E%5Ccap%20C%29)
![\Rightarrow P(E)=0.018+0.012+0.0686](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%29%3D0.018%2B0.012%2B0.0686)
![\Rightarrow P(E)=0.0986](https://tex.z-dn.net/?f=%5CRightarrow%20P%28E%29%3D0.0986)
So, the total probability that the part produced get shipped
![=1-P(E)=1-0.0986=0.9014](https://tex.z-dn.net/?f=%3D1-P%28E%29%3D1-0.0986%3D0.9014)
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)](https://tex.z-dn.net/?f=%3D%5Cleft%28P%28A%29-P%28E%5Ccap%20A%29%5Cright%29%2B%5Cleft%28P%28B%29-P%28E%5Ccap%20B%29%5Cright%29)
![=(0.9-0.018)+(0.03-0.012)](https://tex.z-dn.net/?f=%3D%280.9-0.018%29%2B%280.03-0.012%29)
![=0.9](https://tex.z-dn.net/?f=%3D0.9)
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}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctext%7BProbabilily%20that%20shipped%20part%20is%20%27good%27%7D%7D%7B%5Ctext%7BProbability%20of%20total%20shipped%20parts%7D%7D)
![=\frac{0.9}{0.9014}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.9%7D%7B0.9014%7D)
![=0.9984](https://tex.z-dn.net/?f=%3D0.9984)
(c) The probability that the 'bad' (defective} parts get shipped
=1- the probability that the 'good' parts get shipped
=1-0.9984
=0.0016