
Respuesta :
The probability is 0.0595.
We find this using the formula
[tex]_nC_x(p)^x(1-p)^{n-x} \\ \\=_6C_2{0.7)^2(1-0.7)^{6-2} \\ \\=_6C_2(0.7)^2(0.3)^4 \\ \\=15(0.7)^2(0.3)^4 = 0.0595[/tex]
We find this using the formula
[tex]_nC_x(p)^x(1-p)^{n-x} \\ \\=_6C_2{0.7)^2(1-0.7)^{6-2} \\ \\=_6C_2(0.7)^2(0.3)^4 \\ \\=15(0.7)^2(0.3)^4 = 0.0595[/tex]