Throwing a die three times - check my proof "A teacher rolls a 6-sided die (numbers from 1 to 6) three times. The lowest number will be the grade of the student. 
Calculate the probability of each grade."
My approach is:
Firstly we determine the sample space $\Omega$. So $\Omega:= \{\{\omega_1, \omega_2, \omega_3\},~with~\omega_i \in \{1,...,6\}\}$. The 3-sets $\{\omega_1, \omega_2, \omega_3\}$ are non-ordered and as repitition is allowed we have $\vert\Omega\vert={6+3-1\choose 3}$. 
1.) Let be $A_1$ the set of all 3-sets which at least contain one $1$. We have to consider the cases in which the 3-set contains one $1$, two $1's$ and three $1's$. Hence, $\vert A_1 \vert= {5+2-1 \choose 2}+5+1$. The probabilty of $A_1$ is $\frac{21}{56}= 0.375$.
2.) Let be $A_2$ the set of all 3-sets which at least contain one $2$ and no $1$. We have to consider the cases in which the 3-set contains one $2$, two $2's$ and three $2's$. Hence, $\vert A_2 \vert= {4+2-1 \choose 2}+4+1$. The probabilty of $A_2$ is $\frac{15}{56} \approx 0.268$.
In this manner I would proceed for the remaining 3-sets.
However, the solution comes up with slightly different probabilities as it uses the combinatorial model of "ordered sample" and "repitition allowed"
What is wrong with my approach?
 A: When using counting techniques to calculate probabilities, we may use $Pr(A)=\dfrac{|A|}{|S|}$ where $S$ is the sample space only in the event that every outcome in the sample space is equally likely to occur!
There are two outcomes to a lottery, you either win or you lose, but you don't win the lottery with probability $\frac{1}{2}$.
The outcomes you are counting using stars and bars are not equally likely to occur.  It is far more likely to end up with one die showing a $1$, one showing a $2$, one showing a $3$ than it is to end up with three dice all showing a $1$.

To clarify matters, you may temporarily imagine that the dice being thrown are all different colors.  We recognize that since the results of the dice themselves are independent of one another (as is the standard assumption for thrown dice) that each of the outcomes where "order matters" (or rather where "color matters") will be equally likely to occur.  It is thanks to that that we may calculate the probabilities using these counting techniques.
A: The grade $1$ occurs when $1,2$ or $3$ dice show $1$. 
Only one die showing $1$:
$${3\choose 1}\cdot 5^2$$
Two dice showing $1$ each:
$${3\choose 2}\cdot 5$$
All three dice showing $1$:
$${3\choose 3}$$
Hence:
$$P(1)=\frac{{3\choose 1}5^2+{3\choose 2}5+{3\choose 3}}{6^3}\approx 0.421$$
Similarly:
$$P(2)=\frac{{3\choose 1}4^2+{3\choose 2}4+{3\choose 3}}{6^3}\approx 0.282\\
P(3)=\frac{{3\choose 1}3^2+{3\choose 2}3+{3\choose 3}}{6^3}\approx 0.171\\
P(4)=\frac{{3\choose 1}2^2+{3\choose 2}2+{3\choose 3}}{6^3}\approx 0.088\\
P(5)=\frac{{3\choose 1}1^2+{3\choose 2}1+{3\choose 3}}{6^3}\approx 0.032\\
P(6)=\frac{{3\choose 1}0^2+{3\choose 2}0+{3\choose 3}}{6^3}\approx 0.005\\$$
