\(B=\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}+\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
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