$A$ box contains $9$ tickets numbered $1$ to $9$ both inclusive. If $3$ tickets are drawn from the box one at a time without replacement,then the probability that they are alternatively either {odd,even,odd} or {even,odd,even} is

  • A
    $\frac{5}{17}$
  • B
    $\frac{4}{17}$
  • C
    $\frac{5}{16}$
  • D
    $\frac{5}{18}$

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