$A$ biased die is marked with numbers $2, 4, 8, 16, 32, 32$ on its faces. The probability of getting a face with mark $n$ is $\frac{1}{n}$. If the die is thrown thrice,then the probability that the sum of the numbers obtained is $48$ is:

  • A
    $\frac{7}{2^{11}}$
  • B
    $\frac{7}{2^{12}}$
  • C
    $\frac{3}{2^{10}}$
  • D
    $\frac{13}{2^{12}}$

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