The power set of X will contain 210 elements where each element is subset of A in which
10C0 will contain 0 element.
10C1 will contain 1 element.
10C2 will contain 2 elements.
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10C10 will contain 10 elements.
So total numbers of ways in which we can select two sets with replacement is
And favorable cases would be 10C0⋅10C0+10C1.10C1+……+10C10.10C10=20C10
Hence, Probability would be =22020C10