Records indicate that for parts coming out a hydraulic repair shop at an airplane rework facility, 30/% will have a shaft defect, 15/% will have a bushing defect, and 65/% will be defect-free. Let A="The item has at least one type of defect"; and B="The item has only a shaft defect". Find P (A|B)

P (A|B) = P (A∩B) / P (B) = 100%

Which means that there is 100% probability that the item has at least one type of defect given that the item has only a shaft defect.

Step-by-step explanation:

Conditional probability P (A|B) can be expressed as;

P (A|B) = P (A∩B) / P (B) ... 1

Given;

30% will have a shaft defect,

15% will have a bushing defect,

and 65% will be defect-free

Total probability = 100% = P (shaft or/and bushing defect) + P (defect free)

P (shaft or/and bushing defect) = 100% - P (defect free)

= 100% - 65% = 35%

And

P (shaft or/and bushing defect) = P (shaft def only) + P (bushing def only) + P (shaft and bushing defect)

P (shaft or/and bushing defect) = P (shaft defect) + P (bushing defect) - P (shaft and bushing defect)

Substituting the values we have;

35% = 30% + 15% - P (shaft and bushing defect)

P (shaft and bushing defect) = 45% - 35% = 10%

Let A="The item has at least one type of defect"; and B="The item has only a shaft defect".

P (A) = P (shaft or/and bushing defect) = 35%

P (B) = P (shaft only defect) = 30% - 10% = 20%

P (A∩B) = 20%

Substituting into equation 1

P (A|B) = P (A∩B) / P (B) = 20%/20%

P (A|B) = 1/1 = 100%

Which means that there is 100% probability that the item has at least one type of defect given that the item has only a shaft defect.