If $n(U)=1000$, $n(X)=300$, $n(Y)=400$ and $n(X\cap Y)=200$, then find $n(P(X^{\prime }\cap Y^{\prime }))$.

First, find $n(X\cup Y)$:
$$n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)=300+400-200=500$$
Using De Morgan's Law: $X^{\prime }\cap Y^{\prime }=(X\cup Y{)}^{\prime }$.The number of elements in the complement is:
$$n(X^{\prime }\cap Y^{\prime })=n((X\cup Y{)}^{\prime })=n(U)-n(X\cup Y)=1000-500=500$$
The number of elements in the power set $P(A)$ is $2^{n(A)}$.
$$n(P(X^{\prime }\cap Y^{\prime }))=2^{500}$$