12 coins are tossed 4096 times. The expected number of times that one can get at least 2 heads is

$$\begin{gathered} 12 coins are tossed 4096 way to get no head=1 \\ way to get one head=\frac{1}{2} \\ way to get at least two heads \\ =4096[1-(\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)\left)\right] \\ =4096\left(1-\frac{1}{2^{12}}-\frac{12}{2^{12}}\right)=4096\left(\frac{4096-13}{4096}\right)=4083 \end{gathered}$$