hydraulic landing assemblies coming from an aircraft rework facility are each inspected for defects.
Historical records indicate that 8% have defects in shafts only, 6% have defects in bushing only, and 2% have defects in both shafts and bushing. one of the hydraulic assemblies is selected randomly. What is the probability that the assembly has
a) a bushing defects?
b) a shaft or bushing defect?
c) exactly one of the two types of defects?
d) neither type of defect?
Let event s represent a Shaft defect and event b represent a Brushing defect.
The given probabilities are:
P(s ∩ ¬b) = 0.08
P(b ∩ ¬s) = 0.06
P(b ∩ s) = 0.02
Note: these are mutually exclusive
What you wish to find are:
a: Brushing defect.
P(b) = P(b ∩ ¬s) + P(b ∩ s)
b: Shaft or Brushing.
P(s U b) = P(b ∩ ¬s) + P(s ∩ ¬b) + P(b ∩ s)
c: Shaft and not Brushing OR Brushing and not Shaft
P((s ∩ ¬b) U (b ∩ ¬s)) = P(s ∩ ¬b) + P(b ∩ ¬s)
d: Not Shaft and not Brushing
P(¬s ∩ ¬b) = 1 - P(s U b)
Good post, Graham
Suppose that an assembly operation in a manufacturing plant involves four steps, which
can be performed in any sequence. If the manufacturer wishes to compare the assembly
time for each of the sequences, how many different sequences will be involved in the
experiment?
a) Well, if we look at the information given, we know that 6% of the assemblies have defects in the bushing only. So the probability of a randomly selected assembly having a bushing defect is 6%.
b) To find the probability of a shaft or bushing defect, we need to add the individual probabilities of each type of defect. There is an 8% probability of shaft defects, a 6% probability of bushing defects, and a 2% probability of both types of defects. So, the total probability is 8% + 6% - 2% = 12%.
c) To find the probability of exactly one of the two types of defects, we need to subtract the probability of both types of defects from the sum of the individual probabilities. So, it will be (8% + 6%) - 2% = 12%.
d) Finally, to find the probability of neither type of defect, we need to subtract the sum of the individual probabilities from 100%, as this represents the total population of assemblies without any defects. So, it will be 100% - (8% + 6% - 2%) = 94%.
To solve these probability questions, we will use the concept of the probability of events occurring, denoted as P(event). We will also make use of the concept of mutually exclusive events, which means that the events cannot happen at the same time.
Given:
- Historical records indicate that 8% have defects in shafts only.
- Historical records indicate that 6% have defects in bushing only.
- Historical records indicate that 2% have defects in both shafts and bushing.
a) Probability of having a bushing defect:
To calculate this probability, we need to sum up the percentage of assemblies with bushing defects only and assemblies with both shaft and bushing defects.
P(bushing) = P(bushing only) + P(both defects)
P(bushing) = 6% + 2% = 8%
Therefore, the probability that the assembly has a bushing defect is 8%.
b) Probability of having a shaft or bushing defect:
To calculate this probability, we need to sum up the percentage of assemblies with shaft defects only, assemblies with bushing defects only, and assemblies with both shaft and bushing defects.
P(shaft or bushing) = P(shaft only) + P(bushing only) + P(both defects)
P(shaft or bushing) = 8% + 6% + 2% = 16%
Therefore, the probability that the assembly has a shaft or bushing defect is 16%.
c) Probability of having exactly one of the two types of defects:
To calculate this probability, we need to subtract the probability of having both defects from the probability of having either a shaft or bushing defect.
P(exactly one defect) = P(shaft or bushing) - P(both defects)
P(exactly one defect) = 16% - 2% = 14%
Therefore, the probability that the assembly has exactly one of the two types of defects is 14%.
d) Probability of having neither type of defect:
Since the probability of having neither type of defect includes all the assemblies that do not have any defect, we can calculate it by subtracting the probabilities of all the types of defects from 100% (the total).
P(neither defect) = 100% - P(shaft) - P(bushing) + P(both defects)
P(neither defect) = 100% - 8% - 6% + 2% = 88%
Therefore, the probability that the assembly has neither type of defect is 88%.