Each of the 4 children can be either a boy or a girl

$n\left(S\right)=2^4=16$

$\mathrm{E}= \mathrm{event} \mathrm{of} \mathrm{having} \mathrm{atleast} \mathrm{two} \mathrm{girls}$

$n\left(E\right)={ }^4{C}_2{+}^4{C}_3{+}^4{C}_4=6+4+1=11$

$$\begin{gathered} \text{ Probability of having at least }2\text{ girls }=\frac{\text{ No. of favorable outcomes }}{\text{ Total no. of outcomes }} \\ =\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{11}{16} \end{gathered}$$