$\begin {array} {1 1} (A)\;41 \times \large\frac{2^4}{3^6} & \quad (B)\;\large\frac{2^4}{3^6} \\ (C)\;20 \times \large\frac{2^4}{3^6} & \quad (D)\;None\: of \: these \end {array}$

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The probability of getting atleast 3 in a throw = $ \large\frac{4}{6}=\large\frac{2}{3}$

$ \therefore $ The required probability

$ = 6C_3 \bigg( \large\frac{2}{3} \bigg)^3 \bigg( \large\frac{1}{3} \bigg)^3 + 6C_4 \bigg( \large\frac{2}{3} \bigg)^4 \bigg( \large\frac{1}{3} \bigg)^2 + 6C_5 \bigg( \large\frac{2}{3} \bigg)^5 \large\frac{1}{3} + 6C_6 \bigg( \large\frac{2}{3} \bigg)^6$

$ = \large\frac{41(2^4)}{3^6}$

Hence Ans (A)

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