# Basic Probabilities. Conditional Prob.

1. Apr 7, 2010

### Bacle

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.?

Thanks.

2. Apr 8, 2010

### NoobixCube

At first glance (without going through the math) you have dropped a power of 10 somewhere...

you may want to redo your calculations

3. Apr 8, 2010

### Bacle

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.)?