Picking randomly a number with two digits that can not be divided by $10$

  • $A)$ Find the expectation value and the variance of the number

$\color{gray}{\text{This part is for tommorow:}}$

- $B)$ Find the expectetion value and the variance of the digits multiply- $C)$ Find the probability using Chebyshev's inequality that the number will be $(1.)$ bigger then $85$. $(2.)$ bigger then $25$.

My attempt:

  • $A)$

The numbers with two digits that can not be divided by ten are

$11,\dots,19,21,\dots,29,31,\dots,39,41,\dots,49,51,\dots,59,\dots \dots\dots\dots,99$

$\Omega$ is all this numbers $\uparrow$

$|\Omega|=9\times 9=81$

Let $X$ be a random variable

$$X_1\sim U(11,19)\Longrightarrow E[X_1]=\frac{11+19}{2}=15\Longrightarrow Var[X_1]=\frac{(19-11)^2}{12}=\frac{16}{3}\\ X_2\sim U(21,29)\Longrightarrow E[X_2]=\frac{21+29}{2}=25\Longrightarrow Var[X_2]=\frac{(29-21)^2}{12}=\frac{16}{3}\\ X_3\sim U(31,39)\Longrightarrow E[X_3]=\frac{31+39}{2}=35\Longrightarrow Var[X_3]=\frac{(39-31)^2}{12}=\frac{16}{3}\\ X_4\sim U(41,49)\Longrightarrow E[X_4]=\frac{41+49}{2}=45\Longrightarrow Var[X_4]=\frac{(49-41)^2}{12}=\frac{16}{3}\\ X_5\sim U(51,59)\Longrightarrow E[X_5]=\frac{51+59}{2}=55\Longrightarrow Var[X_5]=\frac{(59-51)^2}{12}=\frac{16}{3}\\ X_6\sim U(61,69)\Longrightarrow E[X_6]=\frac{61+69}{2}=65\Longrightarrow Var[X_6]=\frac{(69-61)^2}{12}=\frac{16}{3}\\ X_7\sim U(71,79)\Longrightarrow E[X_7]=\frac{71+79}{2}=75\Longrightarrow Var[X_7]=\frac{(79-71)^2}{12}=\frac{16}{3}\\ X_8\sim U(81,89)\Longrightarrow E[X_8]=\frac{81+89}{2}=85\Longrightarrow Var[X_8]=\frac{(89-81)^2}{12}=\frac{16}{3}\\ X_9\sim U(91,99)\Longrightarrow E[X_9]=\frac{91+99}{2}=95\Longrightarrow Var[X_9]=\frac{(99-91)^2}{12}=\frac{16}{3}$$

So $E[X]=E[X_1]+\dots+E[X_9]=15+\dots+95=\bbox[yellow]{495}$

S0 $Var[X]=Var[X_1]+\dots+Var[X_9]=\frac{16}{3}\times 9=\bbox[yellow]{48}$

Is my attempt for $A)$ correct?

  • $\begingroup$ Reality Check: How can the expected value be greater than $99$? $\endgroup$ Commented Feb 4, 2016 at 0:21

2 Answers 2


Let $X$ be the first digit, and $Y$ be the last, so $10X+Y$ is our number. $X$ and $Y$ are selected uniformly from $\{1..9\}$.

Then $\mathsf E(10X+Y) = 10\mathsf E(X)+\mathsf E(Y)$ by linearity of expectation.   So we have $$\mathsf E(10X+Y) = 11\times 5 = 55$$

Variance is slightly more involved. $\mathsf {Var}(10X+Y) = 100\mathsf {Var}(X)+\mathsf{Var}(Y)+10\mathsf{Cov}(X,Y)$

However $\mathsf {Cov}(X,Y)= 0$ if the numbers are selected independently, so.

$$\mathsf {Var}(10X+Y) = 101\cdot\dfrac{(9-1+1)^2-1}{12} = 673.\dot{\overline{3}}$$


Let $S=\{0,1,2,3,4,5,6,7,8,9\}$ Then we have $|S|=10$, and the probability that a number is not divisible by 10 can be reduced to a counting problem. We have 9 ways to choose the first digit (since the only number we can't choose is zero) and 9 ways to choose the second digit (again we can't choose 0 or it's divisible by 10). Our entire space of possibilities is $9\cdot 10$ since there are again 9 ways t choose the first digit such that it isn't zero, and 10 ways to choose any remaining digit. Now, if you have the set of all two digit numbers (let's call it $X$) that do not divide 10, each number occurs with equal probability, so by linearity of expectation, your expected value is $(1/81)\cdot \sum_{x\in X} x$ which is 4555. so $(1/81) \cdot 4555$ is your expected value, which turns out to be $\approx $56.2

For variance, you can just do the same thing but $[(1/81)\cdot \sum_{x\in X} x^{2}]$-$56.2^{2}$=$(299565*(1/81)-(56.2)^2)$$\approx 536.5$

  • $\begingroup$ Using the uniform distribution does not work, because the variables in this case are not continuous, and you are considering a sum of random variables which is not exactly what this question is asking (if you had to choose 9 numbers on the other hand, each from a 9 digit interval that you define, then the expected value of their sum would be the expected value you came up with) $\endgroup$
    – mm8511
    Commented Feb 3, 2016 at 22:23
  • 1
    $\begingroup$ The correct sum of $x$ over $x\in X$ is $4455$, giving an expected value of $55$. Or you can just observe that the distribution is symmetric around $x=55$ and therefore it is the mean. The sum of squares appears to be correct, so if you change $56.2$ to $55$ in the calculation of variance I think you get the correct result. $\endgroup$
    – David K
    Commented Feb 3, 2016 at 23:16
  • $\begingroup$ Thanks @DavidK I missed that. $\endgroup$
    – mm8511
    Commented Feb 3, 2016 at 23:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .