Total jurors = \(15\)
Men = \(\frac{2}{3}(15) = 10\)
Women = \(15 - 10 = 5\)

Jury = \(12\)
Men (at-least) = \(\frac{2}{3}(12) = 8\)

Case I: 8M4W
possible selections = \((^{10}C_8)(^5C_4) = 225\)

Case II: 9M3W
possible selections = \((^{10}C_9)(^5C_3) = 100\)

Case III: 10M2W
possible selections = \((^{10}C_{10})(^5C_2) = 10\)

Total possibilities of selecting \(12\) members from \(15\) = \(^{15}C_{12} = 455\)

Required probability = \(\frac{(225 + 100 + 10)}{455} = \frac{335}{455} = \frac{67}{91}\)

Hence, option D