[tex]\mathbb P(X>1)=1-\mathbb P(X\le1)=1-\mathbb P(X=0)-\mathbb P(X=1)[/tex]
The coin is unbiased, so either side has the same probability of [tex]\dfrac12[/tex] of occurring, where
[tex]\mathbb P(X=x)=\dbinom4x\left(\dfrac12\right)^x\left(\dfrac12\right)^{4-x}=\dbinom4x\left(\dfrac12\right)^4=\dfrac1{16}\dbinom4x[/tex]
So the probability of getting heads more than once is
[tex]\mathbb P(X>1)=1-\dfrac1{16}\dbinom40-\dfrac1{16}\dbinom41=\dfrac{11}{16}[/tex]
