Question 875108: 10% of the units in a batch have a common defect. Experience show that 87% of the defective units posses a certain characteristic, while only 3% of the units which do not have this defect posses that characteristic. A unit is examined and found to have the characteristic. What's the conditional probability that the unit has the defect? Thanks.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Events:
C: item possesses a certain characteristic
C': item does not posses a certain characteristic
D: item has a defect
D': item does not have a defect
Probabilities:
P(C): probability that you choose an item that possesses a certain characteristic
P(C'): probability that you choose an item that does not possess a certain characteristic
P(D): probability that you choose an item that is defective
P(D'): probability that you choose an item that is not defective
P(C|D): probability that you choose an item that possesses a certain characteristic given the item has a defect
P(C'|D): probability that you choose an item that does not possess a certain characteristic given the item has a defect
P(C|D'): probability that you choose an item that possesses a certain characteristic given the item does not have a defect
P(C'|D'): probability that you choose an item that does not possess a certain characteristic given the item does not have a defect
P(D|C): probability that you choose an item that is defective given the item has that certain characteristic
P(D'|C): probability that you choose an item that is not defective given the item has that certain characteristic
P(D|C'): probability that you choose an item that is defective given the item does not have a certain characteristic
P(D'|C'): probability that you choose an item that is not defective given the item does not have a certain characteristic
-----------------------------------------------------------------
Given Probabilities
"10% of the units in a batch have a common defect"
P(D) = 0.10
P(D') = 0.90 (1 - 0.10 = 0.90)
"Experience show that 87% of the defective units posses a certain characteristic"
P(C|D) = 0.87 (given a defective unit, 87% of them will have this characteristic)
"only 3% of the units which do not have this defect posses that characteristic"
P(C|D') = 0.03 (given a non-defective unit, 3% of them will have this characteristic)
-------------------------------------------------------
"A unit is examined and found to have the characteristic. What's the conditional probability that the unit has the defect?" so we want to find P(D|C). This is the probability of event D happening given event C has happened.
We want to find P(D|C)
Use Baye's Theorem to get
P(D|C) = (P(C|D)*P(D))/(P(C))
Then use the Law of Total Probability to break up P(C) into P(C|D)*P(D)+P(C|D')*P(D')
We will now have this equation
P(D|C) = (P(C|D)*P(D))/(P(C|D)*P(D)+P(C|D')*P(D'))
Plug in the values: P(D) = 0.10, P(D') = 0.90, P(C|D) = 0.87, P(C|D') = 0.03 and evaluate
P(D|C) = (P(C|D)*P(D))/(P(C|D)*P(D)+P(C|D')*P(D'))
P(D|C) = (0.87*0.10)/(0.87*0.10+0.03*0.90)
P(D|C) = 0.76315789473684
|
|
|
