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.