Question

Barbara has a jar of coins, some Canadian, some U.S., and similarly, some quarters, some dimes. [There is no other nationality of coin, nor any other value of coin.] 70% are Canadian coins. Given that it is a Canadian coin, there is a 40% chance it is a quarter. Given that it is not a Canadian coin, there is a 40% chance it is a quarter. Are nationality and value independent? Or is impossible to tell? Please explain how you know.

Answer 1

We have given

p(canadian)=0.7

p(Quarter/Canadian)=0.4 p(Quarter and Canadian)/p(Canadian)=0.4 p(Quarter and Canadian)=0.4*0.7=0.28

p(Quarter and Canadian)=0.28 ----------------------- (1)

We want to find  p(Quarter)*p(Canadian)

lets find p(Quarter)

As per bays rule

p(Quarter) =  p(Quarter / Canadian)*p(Canadian) + p(Quarter / Non Canadian)*p(Non Canadian)

{we know p(Quarter/Non Canadian)=0.4 ; p(Quarter/Canadian)=0.4 ; p(canadian)=0.7; p(Non canadian)=0.3}

=0.4*0.7+0.4*0.3

=0.4

p(Quarter)*p(Canadian)=0.4*0.7=0.28

p(Quarter)*p(Canadian)=0.28 --------------------------------------------(2)

We can see   p(Quarter and Canadian)=0.28 ----------------------- (1)

p(Quarter)*p(Canadian)=0.28 --------------------------------------------(2)

therefore  p(Quarter and Canadian)=p(Quarter)*p(Canadian) nationality and value independent

Conclusion:

Yes nationality and value independent.

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