Dtyjlui

Say a fair dice is thrown $n$ times. Showing using the Chebyshev Inequality that the probability that the number of sixes thrown lies between $frac{1}{6}n-sqrt{n}$ and $frac{1}{6}n+sqrt{n}$ is at least $frac{31}{36}$.

Idea:

The at least leads me to believe that I be doing it over the complement of ${frac{1}{6}n-sqrt{n}< X<frac{1}{6}n+sqrt{n}}={-sqrt{n}< X-frac{1}{6}n<sqrt{n}}$ and note $mathbb E[X]=frac{n}{6}$, where $X$ is number of sixes thrown

This looks similar to Chebyshev.
So, for $epsilon > 0$

$P(|X-frac{1}{6}n|< sqrt{n})=1-P(|X-frac{1}{6}n|geq sqrt{n})leq1-frac{operatorname{Var}{X}}{epsilon^2}iff P(|X-frac{1}{6}n|geqsqrt{n})leqfrac{operatorname{Var}{X}}{epsilon^2}$

But how can I calculate $operatorname{Var}{X}$? Do I have to explicitly define the Random Variable and then find an appropriate density funtion? All I have is $operatorname{Var}{X}=mathbb E[(X-mathbb E[X])^2]=int X-frac{1}{6}noperatorname{dP}$

Any help is greatly appreciated.

You're sort of on the right track, but your inequality in $1-Pleft(|X-frac{1}{6}n|geq sqrt{n}right)leq1-frac{operatorname{Var}{X}}{epsilon^2}$ is the wrong way round ( it should be $1-Pleft(|X-frac{1}{6}n|geq sqrt{n}right)geq1-frac{operatorname{Var}{X}}{epsilon^2}$ ), you haven't identified what $epsilon$ is, and a power of 2 is missing from your integral for the variance. The latter should be $int left( X - frac{1}{6} n right)^2 dP$ .

I presume the form of Chebyshev's inequality you're using is $P(|X-frac{1}{6}n|geq epsilon)leqfrac{operatorname{Var}{X}}{epsilon^2}$ , in which case your $epsilon$ is just $sqrt{n}$ , and your inequality becomes $P(|X-frac{1}{6}n|geq sqrt{n})leqfrac{operatorname{Var}{X}}{n}$

You could evaluate the integral for the variance by working out what the distribution $F_X$ of $X$ is (Hint: $F_Xleft(jright) = Pleft(X=jright)$ is the probability of getting $j$ sixes with $n$
independent throws of a fair die), but there's also a simpler way of calculating it.

If $X_i$ is the number of sixes you get on the $i^mbox{th}$ throw, then $Pleft( X_i = 1 right) = frac{1}{6}$ , $Pleft( X_i = 0 right) = frac{5}{6}$ , and $X_1, X_2, dots , X_n$ are independent, identically distributed random variables with $X = X_1 + X_2 + dots + X_n$. Now there's a theorem which tells us that the variance of a sum of $n$ independent identically distributed random variables is just $n$ times the common variance of the summands. That is, $mbox{Var}left(Xright) = n mbox{Var}left(X_1right)$ , so you can prove your result just by calculating the variance of the simple two-valued random variable $X_1$ .

Elaboration of hint about $F_X$:

Since $X$ can only take on one of the values $0, 1, dots , n$ , the sample space (call it $Omega$) can be partitioned into a union of the disjoint events $ E_j = left{ omega in Omega | Xleft(omegaright) = j right} mbox{ for } j=0, 1, dots , n $ . The integral $int left( X - frac{1}{6} n right)^2 dP$ can then be written as $int_{bigcup_{j=0}^n E_j}left( X - frac{1}{6} n right)^2 dP = sum_{j=0}^n int_{E_j}left( X - frac{1}{6} n right)^2 dP $ . Since $X$ has the fixed value $j$ everywhere in $E_j$ , then $int_{E_j}left( X - frac{1}{6} n right)^2 dP = $ $ left( j- frac{1}{6} n right)^2 int_{E_j} dP = left( j- frac{1}{6} n right)^2 Pleft(E_jright) = left( j- frac{1}{6} n right)^2 F_Xleft(jright)$ . So $Varleft(Xright) = sum_{j=0}^n left( j- frac{1}{6} n right)^2 F_Xleft(jright)$ .

As callculus noted in his answer, $X$ is $ n, frac{1}{6}$-binomially distributed, which gives you the expression for $F_Xleft(jright)$ as a function of $n$ and $j$ . If you don't know this expression, you will find it (as well as its variance!) in any good text on elementary probability theory (such as Volume 1 of William Feller's classic, An Introduction to Probability Theory and Its Applications—3rd Edition—where you will find the material on pp.147-8 and p.230) .

Hint: The number of sixes are distributed as $Xsim Binleft( n, frac16right)$. The variance of $X$ is well known. Then indeed you can use the inequality

$$P(|X-frac{1}{6}n|< sqrt{n})geq 1-frac{operatorname{Var}{X}}{epsilon^2}$$

I think you know what to plug in for $epsilon^2$.