probability of finding a rotten egg in A carton is 0.125
probability of finding a rotten egg in B carton is 0.2342
probability of finding a rotten egg in C carton is 0.5337
each carton has 12 eggs.
John found that one egg is rotten, what is the conditional probability that it was from A carton?
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John will take one carton
X i.e. probability of being A carton
Y i.e. probability of being B carton
Z i.e. probability of being C carton
p(x)=p(y)=p(z)=(1/3)
K i.e. probability of rotten egg
p(k/x)=.0125=(1/12)
p(k/y)=.234=(2/12)
p(k/z)=.533=(4/12)
now p(x/k)=[{p(x)p(k/x)}÷{p(x)p(k/x)+p(y)p(k/y)+p(z)p(k/z)}]
=ans
X i.e. probability of being A carton
Y i.e. probability of being B carton
Z i.e. probability of being C carton
p(x)=p(y)=p(z)=(1/3)
K i.e. probability of rotten egg
p(k/x)=.0125=(1/12)
p(k/y)=.234=(2/12)
p(k/z)=.533=(4/12)
now p(x/k)=[{p(x)p(k/x)}÷{p(x)p(k/x)+p(y)p(k/y)+p(z)p(k/z)}]
=ans
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