Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.)?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook