A 6-sided fair die is rolled 24 times. Let $X$ is the sum of all results. Find the probability $P(X \geq 86)$. How can I proceed with this problem? 
I started by finding z score in P:
$P(X ≥ 86) = P(Z > (86-24)/σ)$. Is that correct way?
 A: Yes you can approximate the distribution of $Z = \frac{X-\mathbb{E}[X]}{\sqrt{Var(X)}}$ with a $N(0,1)$ distribution, but you would need to calculate both $\mathbb{E}[X]$ and $Var(X)$ in order to do this. It is useful to note, that since $X$ can be written as a sum of i.i.d. variables, we can calculate
$$\mathbb{E}[X] = \sum_{i=1}^{24} \mathbb{E}[X_i] = 24 \mathbb{E}[X_1]$$
where $\mathbb{E}[X_1]$ is expected value of one dice roll, and similarly by properties of independent variables we can calculate
$$Var(X) = 24Var(X_1)$$
where $Var(X_1)$ is the variance of a single dice roll. Once you have computed $\mathbb{E}[X_1]$ and $Var(X_1)$ you should be able to solve the problem.
A: Since you asked only for hints, here it is:
First step is to find out the sample space.
Here are set of events:
(1,1,1... upto 24 times)
(2,2,2... upto 24 times)
_ _ _ _ _ ... 24 places
In each place you can put one out of 6 digits 
Therefore sample space is $6^{24}$
Now you have to see the digit 86.
Think of it as you have 86 envelops and 24 boxes.
You can put at most 6 envelopes in one box and at least one envelope in one box.
There is basically a ready made formulae available for this.
The no. of positive integral solutions of
$x_1 + x_2$ + ... upto $x_r$ = n is $^{n-1}C_{r-1}$ 
In this case, n= 86 and r=24
By this way you find the total no. of cases when the sum of outcomes equal to 86  
