6 unbaised coins are tossed. The probability of getting atleast 4 heads is 
 

Given  n=6 and probability of getting head $\mathrm{p}=\frac{1}{2}$ and getting tail $\mathrm{q}=\frac{1}{2}$.

 Let X denotes the no. of heads.

Now $\mathrm{P} (X\ge 4)=\mathrm{P}(X=4)+\mathrm{P}(X=5)+\mathrm{P}(X=6)$

                          $$\begin{gathered} ={}^6{\mathrm{C}}_4 {\left(\frac{1}{2}\right)}^4 {\left(\frac{1}{2}\right)}^2+{}^6{\mathrm{C}}_5 {\left(\frac{1}{2}\right)}^5 \left(\frac{1}{2}\right)+{}^6{\mathrm{C}}_6 {\left(\frac{1}{2}\right)}^6 \\ = \left(15+6+1\right) {\left(\frac{1}{2}\right)}^6 \\ =\frac{22}{64} \\ =\frac{11}{32} \end{gathered}$$