If $7$ different balls are distributed among $4$ different boxes,then the probability that the first box contains $3$ balls is

  • A
    $\frac{35}{128}\left(\frac{3}{4}\right)^3$
  • B
    $\frac{35}{64}\left(\frac{3}{4}\right)^4$
  • C
    $\frac{7}{8}\left(\frac{3}{4}\right)^7$
  • D
    $\frac{5}{16}\left(\frac{3}{4}\right)^5$

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