The power set of X will contain $2^{10}$ elements where each element is subset of A in which 
${}^{10}{C}_0$ will contain $0$  element.
${}^{10}{C}_1$ will contain $1$ element.
${}^{10}{C}_2$ will contain $2$ elements.
$⋮$

$⋮$
${}^{10}{C}_{10}$     will contain $10$ elements.
So total numbers of ways in which we can select two sets with replacement is 
And favorable cases would be ${}^{10}{C}_0{\cdot }^{10}{C}_0{+}^{10}{C}_1{.}^{10}{C}_1+ \dots \dots {+}^{10}{C}_{10}{.}^{10}{C}_{10}{=}^{20}{C}_{10}$
Hence, Probability would be  $= \frac{{}^{20}{C}_{10}}{2^{20}}$