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However of the 12, [highlight]"atleast "[/highlight] 2/3 should be men which means we should have atleast 8 or more men. This leads to the following selections
8m 4w 9m 3w 10m 2w
so the prob is (10c8*5c4 + 10c9*5c3 + 10c10*5c2)/15c12 ===> 67/91
now if we need to have have at least 2/3 men in 12 member jury, that means we need to have at least 8 men,
Now the \(P (of having at least 2/3 men in 12 member jury)= 1 - P ( not having 2/3 men in the jury)\)
now, there is only one possiblity when we cant have 8 men in the 12 member jury i.e when all the women are selected. ( 5 women and 7 men)
So the \(probablity of having 5 women and 7 men = \frac{5C_5 * 10C_7}{15C_12}\) \(15C_12\) = total ways of choosing 12 members out of 15 \(P (of having at least 2/3 men in 12 member jury) = 1 - \frac{10*9*8*3*2}{3*2*15*14*13} = 1- \frac{24}{91}\) \(=\frac{67}{91}\)
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