Ten stones all have distinct weights.

They are randomly labeled A through J.

The stone labeled B is weighed against all the other stones except J, and found to be heavier than all of them except A.

The probability that J is heavier than B is given as $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers.

What is $a+b$ ?

The answer is 6.

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There's a lot of "noise" in this problem. To clear the noise, note that $A, J > B > C \mbox{ through } I$ just means " $J$ is the heaviest or the second heaviest".

Out of 10 stones, what is the probability that we would choose either the heaviest or the second heaviest stone to be our "J"?

Answer: $\frac{2}{10} = \frac{1}{5}$

so $a+b = 1+5 = \boxed{6}$ .