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Basic Probabilities. Conditional Prob.

  1. Apr 7, 2010 #1
    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:
    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.

    P(A|B) is conditional probability of B, given A. '/\' means

    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

    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.?

  2. jcsd
  3. Apr 8, 2010 #2
    At first glance (without going through the math) you have dropped a power of 10 somewhere...

    you may want to redo your calculations
  4. Apr 8, 2010 #3
    Thanks, Noobix.
    You mean because Z should be much more likely to have broken the plate than either
    X or Y (which would not be the case if P(Z|Br) was 0.057.)?
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