- #1
Bacle
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Hi, everyone : I have the following problem:
We have 3 dishwashers X,Y,Z, with the conditions:
1) X washes 40% of dishes, and breaks
1% of the dishes s/he washes.
2)Y washes 30% of the dishes, breaks 1%
3)Z washes 30% of the dishes and breaks 3%.
Question: If a dish is broken: what is the probability that Z broke
the dish.?.
My work:
Events:
E1)Br means "Broke the dish",
E2)X (equiv. Y,Z) means "X ( Equiv. Y,Z) washed the dish.".
E3) (Br|X) means event of dish breaking when X is washing.
Notation:
P(A|B) is conditional probability of B, given A. '/\' means
intersection.
We have : P(Br|X)= 0.01 , P(Br|Y)=0.01 and P(Br|Z)=0.03
P(X)=0.4 , P(Y)=0.3 , P(Z)=0.3
We want to find P(Z|Br), which is equal to P(Z/\Br)/P(Br) , by def. of conditional
probability.
1) First, we find P(Br)=P( (Br/\X)\/(Br/\Y)\/(Br/\Z) )=
P(Br|X)P(X)+P(Br|Y)P(Y)+ p(Br|Z)P(Z)= 0.16
(side question: how do we know that any assignment of probabilities here will
give us P(Br)< =1 ? )
2) P(Z/\Br) =P(Z)P(Br|Z) = (0.3)(0.03)=0.009
3) Using 1,2 above, we get P(Br|Z)= 0.009/0.16= 9/160= 0.05625
But the book has 0.57 as a solution. Could the book have made a mistake.?
Would anyone please check.?
Thanks.
We have 3 dishwashers X,Y,Z, with the conditions:
1) X washes 40% of dishes, and breaks
1% of the dishes s/he washes.
2)Y washes 30% of the dishes, breaks 1%
3)Z washes 30% of the dishes and breaks 3%.
Question: If a dish is broken: what is the probability that Z broke
the dish.?.
My work:
Events:
E1)Br means "Broke the dish",
E2)X (equiv. Y,Z) means "X ( Equiv. Y,Z) washed the dish.".
E3) (Br|X) means event of dish breaking when X is washing.
Notation:
P(A|B) is conditional probability of B, given A. '/\' means
intersection.
We have : P(Br|X)= 0.01 , P(Br|Y)=0.01 and P(Br|Z)=0.03
P(X)=0.4 , P(Y)=0.3 , P(Z)=0.3
We want to find P(Z|Br), which is equal to P(Z/\Br)/P(Br) , by def. of conditional
probability.
1) First, we find P(Br)=P( (Br/\X)\/(Br/\Y)\/(Br/\Z) )=
P(Br|X)P(X)+P(Br|Y)P(Y)+ p(Br|Z)P(Z)= 0.16
(side question: how do we know that any assignment of probabilities here will
give us P(Br)< =1 ? )
2) P(Z/\Br) =P(Z)P(Br|Z) = (0.3)(0.03)=0.009
3) Using 1,2 above, we get P(Br|Z)= 0.009/0.16= 9/160= 0.05625
But the book has 0.57 as a solution. Could the book have made a mistake.?
Would anyone please check.?
Thanks.