Question

Suppose two random variables, X and Y are independent, which statement is false? a) P(X∣Y)=P(X) b) P(X∪Y)=P(X)+P(Y) c) P(X∩Y)=P(X)⋅P(Y) d) Cov(X,Y)=0

Answer 1

B is false, since by the inclusion/exclusion principle,

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$

By independence, we have $P(X\cap Y)=P(X)P(Y)$, which is zero if either of $P(X)$ or $P(Y)$ is 0, which isn't guaranteed.