Respuesta :
Knowledge on the definition and notation of conditional probability is also required.
P(A|B) means probability that event A is a success given that B has happened.
Definition: P(A|B)=P(AnB)/P(B).
"n" is used to mean ∩ or the intersection symbol in sets
Bayes theorem is the key to the solution.
Here we are given
Let d=event that a unit is defective
Total daily production = 2000
P(i)=800/2000=0.4 P(D|i)=0.07
P(ii)=500/2000=0.25 P(D|ii)=0.05
P(iii)=700/2000=0.35 P(D|iii)=0.06
[total =sum P()=1.0]
We need to find P(ii|D), i.e. probability that unit is produced from plant ii given that it is defective.
overall probability of a defective unit
P(D)=P(D|i)*P(i) + P(D|ii)*P(ii) + P(D|iii)*P(iii) [law of total probability]
=0.4*0.07+0.25*0.05+0.35*0.06
=0.0615
Apply Baye's theorem:
P(ii|D)=P(ii n D)/P(D) [ definition of conditional probability]
=P(D|ii)*P(ii) / P(D) [ application of Baye's theorem, or def. of cond. prob.]
=0.05*0.25/0.0614
=0.2036
Comment on result:
The proportion of units produced in ii is 0.25. However, the defective rate is the lowest of the three plants, hence it is reasonable that the probability that ii produced the defective unit is lower than 0.25, like 0.2036.
Logical glance like this gives a quick check of answers.
P(A|B) means probability that event A is a success given that B has happened.
Definition: P(A|B)=P(AnB)/P(B).
"n" is used to mean ∩ or the intersection symbol in sets
Bayes theorem is the key to the solution.
Here we are given
Let d=event that a unit is defective
Total daily production = 2000
P(i)=800/2000=0.4 P(D|i)=0.07
P(ii)=500/2000=0.25 P(D|ii)=0.05
P(iii)=700/2000=0.35 P(D|iii)=0.06
[total =sum P()=1.0]
We need to find P(ii|D), i.e. probability that unit is produced from plant ii given that it is defective.
overall probability of a defective unit
P(D)=P(D|i)*P(i) + P(D|ii)*P(ii) + P(D|iii)*P(iii) [law of total probability]
=0.4*0.07+0.25*0.05+0.35*0.06
=0.0615
Apply Baye's theorem:
P(ii|D)=P(ii n D)/P(D) [ definition of conditional probability]
=P(D|ii)*P(ii) / P(D) [ application of Baye's theorem, or def. of cond. prob.]
=0.05*0.25/0.0614
=0.2036
Comment on result:
The proportion of units produced in ii is 0.25. However, the defective rate is the lowest of the three plants, hence it is reasonable that the probability that ii produced the defective unit is lower than 0.25, like 0.2036.
Logical glance like this gives a quick check of answers.