Finding probability using Chebyshev's Theorem

Let $X_1,...,X_n$ be i.i.d random variables. $X_i$ ~ $Unif(-1,1)$

Let $Y_1,...,Y_n$ be i.i.d random variables. $~~$$Y_i ~ Unif(-1,1) Let us define a new random variable Z_i. Z_i=1 if (X_i,Y_i) lies within the unit-disk. And Z_i=0 if (X_i,Y_i) lies outside the unit disk. . What is the mean and variance of \bar{Z_n}? What about 4\bar{Z_n}? . Is 4\bar{Z_n} close to \pi? Find P(|4\bar{Z_9}-\pi|>.01). Use Chebyshev's theorem to find bound on this probability. What about P(|4\bar{Z}_{100}-\pi|>.01)? . Instead of finding a bound as in the last example, find the P(|4\bar{Z_9}-\pi|>.01) approximately using Central Limit Theorem. I am having trouble using Chebyshev's theorem. I think my Var(4\bar{Z}_{9}) might be wrong but unsure why.$$Z_i \sim Bern(\frac{\pi}{4})E(\bar{Z_n})=\dfrac{\pi}{4} ,~ Var(\bar{Z_n})=(\dfrac{\pi}{4n}-\dfrac{\pi^2}{16n})E(4\bar{Z_n})=\pi ,~ Var(4\bar{Z_n})=\dfrac{\pi}{n}(4-\pi)$$Since P(|X-E(x)|>k)\leq\dfrac{Var(X)}{k^2}$$P(|4\bar{Z_9}-\pi|>.01)\leq \dfrac{(4\pi - \pi^2)}{9(.01^2)}$$But this number is clearly incorrect. Where did I go wrong? • Why is it clearly incorrect? It looks correct. – E-A Sep 3 '18 at 23:13 • the bound evaluates to almost 3000 @E-A. Does this matter? – 1337 Sep 3 '18 at 23:16 • No; it would be an issue if it were, say, negative. This is after all a "bound"; it does not need to be a good one. This bound will get better as n goes to infinity. (i.e. when you put n instead of 9) – E-A Sep 3 '18 at 23:19 • Am I correct in saying$Z_i \sim Bern(\dfrac{\pi}{4})$@E-A? – 1337 Sep 3 '18 at 23:22 • As always I would like to point out that independence assumptions are important in probability. This question has no answer unless$X_i$'s are independent of$Y_i\$'s. – Kavi Rama Murthy Sep 3 '18 at 23:34