Let $(X,Y)$ be two dimensional random variable such that $E(X)=E(Y)=3$ & $var(X)=var(Y)=1$ & $cov(X,Y)=\dfrac{1}{2}$. Then , $P(|X-Y|>6)$ is :
(a) less than $\dfrac{1}{6}$
(b) equal to $\dfrac{1}{2}$
(c) equal to $\dfrac{1}{3}$
(d) greater than $\dfrac{1}{2}$.
$\mathcal My$ $\mathcal Attempt :$
First suppose that, $Z=X-Y$. Then from the given conditions we compute $E(XY)=\dfrac{19}{2}$ & $E\bigl(X^{2}\bigr)=E\bigl(Y^{2}\bigr)=10$. Then compute mean of $Z\bigl(i.e. m_{z} \bigr)$ & variance $\bigl(\sigma_{z}^{2}\bigr)$. We get, $m_{z}=0$ & $\sigma_{z}^{2}=1$.
Thus we get, $Z$ is $N(0,1)$ variate. We have to find $P\bigl(|Z|>6\bigr)$.
Now from the distribution of $Z$ how we can find out the required probability or we conclude any one of the given four options ?
I also tried through $Tchebycheff's$ inequality but I can not apply it.
The answer is clearly (a).
You can discard the "equals" right away - these is not enough information for such a strong conclusion.
You can discard (d) because a random variable with mean $0$ and variance $1$ cannot be so far away from $0$ half the time.
You can use Chebyshev Inequality to actually prove (a):
$$ P(|Z|>6\sigma_Z)\le \frac{1}{6^2} $$
since $\sigma_Z=1$, we get
$$ P(|Z|>6)\le \frac{1}{36} < \frac{1}{6} $$