Let $(X, Y)$ be a 2D random variable such that $E(X)=E(Y)=3$ and variance of both $X$ and $Y$ is $1$ . Covariance between $X$ and $Y$ is $1/2$. Then what is $P(|X-Y| >6)$?
I think chebyshev's inequality for multidimensional random variables can be applied, but I can't reach anywhere with that.
Note that $$ P(|X-Y|>6)\leq\frac{ E(X-Y)^{2}}{6^2}. $$ To compute $E(X-Y)^{2}$, observe that $$ E(X-Y)^{2}=EX^2-2EXY+EY^2\tag{1}\label{1}. $$ Also $$ 1=\text{Var}(X)=EX^2-(EX)^2=EX^2-3^2 $$ and similarly for $\text{Var}(Y)$. Lastly $$ 0.5=\text{Cov}(X,Y)=EXY-EX\cdot EY=EXY-9. $$ You have everything to plug into $\eqref{1}$.