If $12$ identical balls are to be placed randomly in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls is

  • A

    $\frac{4}{{19}}$

  • B

    $\frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$

  • C

    $\frac{{\left( {428} \right){}^{12}{C_3}}}{{{3^{11}}}}$

  • D

    $\frac{5}{{19}}$

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