Solution:
We know that, $\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{P}){^\text{r}.(\text{q})6^{\text{n}-\text{r}}}$
Here, $\text{n}=10,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
and $\text{r}\geq8\text{ i.e,}\text{ r}=8,9,10$
$\Rightarrow\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=8)+\text{P}(\text{r}=9)+\text{P}(\text{r}=10) $
$={^{10}}\text{C}_8\Big(\frac{1}{2}\Big)^8\Big(\frac{1}{2}\Big)^{10-8}+{^{10}}\text{C}_9\Big(\frac{1}{2}\Big)^9\Big(\frac{1}{2}\Big)^{10-9}+{^{10}}\text{C}_{10}\Big(\frac{1}{2}\Big)^{10}\Big(\frac{1}{2}\Big)^{10-10}$
$=\Big(\frac{10!}{8!2!}+\frac{10!}{9!1!}+1\Big)\Big(\frac{1}{2}\Big)^{10}$
$=[45+10+1]\Big(\frac{1}{2}\Big)^{10}$
$=56\Big(\frac{1}{2}\Big)^{10}=\frac{7}{128}$
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