Probability- 6 digit number that is built from the numbers 2,5,6,9

Question

A random 6 digit number is picked that is built only from the numbers $2, 5, 6, 9.$

1. What is the probability that the number can be divided by $3$?

2. What is the probability that the number can be divided by $6$?

3. What is the expected value of the number?

4. What is the variance of the number?

I am stuck the 3rd question... I know how to solve the $4$th part with the next equation- variance and expected value connection but have no idea how to even find the expected value..

And about the variance- I have $2$ possible answers I got but I do not know what is correct-

a. $6.313\times10^{10)}$, my way of getting this answer is- $E(x^2)=36.5\dots,E^2(x)=30.25\dots,$ so then $E(x^2)-E^2(x)=6.313\times 10^{10}=V(x).$

b. $3.13\times10^{11}$

Which variance is correct?

Answer 1

The expect value of the thousand digit is $\frac{2+5+6+9}{4}=\frac{22}{4}$

Similar for all other digits.

So the expect value is $100000\cdot\frac{22}{4}+10000\cdot\frac{22}{4}+1000\cdot\frac{22}{4}+100\cdot\frac{22}{4}+10\cdot\frac{22}{4}+\cdot\frac{22}{4}=611110.5$

Answer 2

HINT: Since we don't have $0$ in our number list, all these numbers have the same probability of being in ones, tens, hundreds or thousands. Also by symmetry, each number has the same number of occurrences in total. So actually you can consider just the numbers $222222,555555,666666,999999$ to find the expectation.

Answer 3

The number is the weighted sum of the six digits with weights $1,10 ,100 \cdots$.

If $N= a X + b Y$ and $X,Y$ are random independent variables with the same distribution (i.i.d), then if each of them has mean $\mu$ and variance $\sigma^2$ then you know that the mean and variance of $N$ are given by ...

refer to this and to this Wikipedia articles.