$\begin {array} {1 1} (A)\;\large\frac{3}{56} & \quad (B)\;\large\frac{1}{56} \\ (C)\;\large\frac{12}{56} & \quad (D)\\large\frac{14}{56} \end {array}$

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Three possible cases are

$Ist \: draw\: \: \: \: \: IInd \: draw\: \: \: \: \: IIIrd\: draw$

$ Red\: \: \: \: \: \: \: \: \: \: \: \: \: \: Non\: Red\: \: \: \: \: \: \: \: Red$

$Non\: Red\: \: \: \: \: \: \: \: \: \: \: Red\:\: \: \: \: \: \: \: \: \: \: Red$

$ Non\: Red\: \: \: \: \:\: \: \: \: Non\: Red\: \: \: \: \: Red$

$ \therefore $ Required probability = $ \large\frac{2}{8} \times \large\frac{6}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{2}{7} \times \large\frac{1}{6}+ \large\frac{6}{8} \times \large\frac{5}{7} \times \large\frac{2}{6}$

$ = \large\frac{14}{56}$

Ans : (D)

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