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(D) The total number of elements in the power set $P(X)$ is $2^{10}$.Since $A$ and $B$ are chosen with replacement,the total number of ways to choose

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$\frac{^{20}C_{10}}{2^{20}}$

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(D) The total number of elements in the power set $P(X)$ is $2^{10}$.Since $A$ and $B$ are chosen with replacement,the total number of ways to choose the pair $(A, B)$ is $(2^{10}) 	imes (2^{10}) = 2^{20}$.Let $n(A) = k$ and $n(B) = k$,where $k$ can range from $0$ to $10$.The number of ways to choose a subset with $k$ elements from a set of $10$ elements is $^{10}C_k$.Thus,the number of ways to choose $A$ and $B$ such that $n(A) = n(B) = k$ is $(^{10}C_k) 	imes (^{10}C_k) = (^{10}C_k)^2$.The total number of favorable outcomes is $\sum_{k=0}^{10} (^{10}C_k)^2$.Using the identity $\sum_{k=0}^{n} (^{n}C_k)^2 = ^{2n}C_n$,we get $\sum_{k=0}^{10} (^{10}C_k)^2 = ^{20}C_{10}$.Therefore,the required probability is $\frac{^{20}C_{10}}{2^{20}}$.

Let $X$ be a set containing $10$ elements and $P(X)$ be its power set. If $A$ and $B$ are picked up at random from $P(X)$ with replacement,then the probability that $A$ and $B$ have an equal number of elements is:

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