dimitri92 wrote:

Can someone try to solve this?

2/3 men out of 15 = 10 men

1/3 women = 5

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

\(P (of having at least 2/3 men in 12 member jury) = 1 - \frac{10*9*8*3*2}{3*2*15*14*13} = 1- \frac{24}{91}\) \(=\frac{67}{91}\)