Let $\mathrm{X}$ is the random variable for getting a head Here, $n=10, r \geq 8$
$$
\Rightarrow \mathrm{r}=8,9,10, \mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{1}{2}
$$
We know that,
$\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}$
$\mathrm{P}$ (atleast 8 heads)
$={ }^{10} \mathrm{C}_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{10-8}+{ }^{10} \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{10-9}$
$+{ }^{10} \mathrm{C}_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10}$
$=\frac{10 !}{8 ! 2 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{9 ! 1 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{0 ! 10 !}\left(\frac{1}{2}\right)^{10}$
$=\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right]$
$=\left(\frac{1}{2}\right)^{10} .56=\frac{1}{2^7 \cdot 2^3} \cdot 56=\frac{7}{128}$