Answer
$p(x\geq3)=.945$
Work Step by Step
$q=1-p=1-.5=.5$
$p(x)=\begin{pmatrix} n \\ x \end{pmatrix}p^xq^{n-x}$
$p(x=0)=\begin{pmatrix} 10 \\ 0 \end{pmatrix}(.5)^0(.5)^{10-0}=\frac{10!}{0!(10-0)!}(.5)^{10}=\frac{10!}{10!}(.5)^{10}=(.5)^{10}$
$p(x=1)=\begin{pmatrix} 10 \\ 1 \end{pmatrix}(.5)^1(.5)^{10-1}=\frac{10!}{1!(10-1)!}(.5)^{10}=\frac{10(9!)}{9!(1!)}(.5)^{10}=10(.5)^{10}$
$p(x=2)=\begin{pmatrix} 10 \\ 2 \end{pmatrix}(.5)^2(.5)^{10-2}=\frac{10!}{2!(10-2)!}(.5)^{10}=\frac{10(9)(8!)}{2!(8!)}(.5)^{10}=45(.5)^{10}$
$p(x\geq3)=1-p(x\lt3)=1-p(x=0)-p(x=1)-p(x=2)=1-(.5)^{10}-10(.5)^{10}-45(.5)^{10}=1-56(.5)^{10}=.945$