Roll a a die three times Draw probability mass function for the following random variables.
I have tried $P(X=1)=(6/6)^3-(5/6)^3$ which gives me the minimum value rather than the maximum value. 
For $P(X=6)=1/6^3$.
 A: Hint:
For the maximum to be $n$ 


*

*you want all $3$ dice to be less than or equal to $n$ 

*but you do not want all $3$ dice to be strictly less than $n$
A: Let $x_1$, $x_2$ and $x_3$ be the three rolls. Then for $1\leq k\leq 6$,
$$P(\max(x_1,x_2,x_3)\leq k)=P(x_1\leq k)\cdot P(x_2\leq k) \cdot P(x_3\leq k)=\left(\frac{k}{6}\right)^3.$$
Hence 
$$\begin{align}P(\max(x_1,x_2,x_3)=k)&=P(\max(x_1,x_2,x_3)\leq k)-P(\max(x_1,x_2,x_3)\leq k-1)\\&=\left(\frac{k}{6}\right)^3-\left(\frac{k-1}{6}\right)^3=\frac{k^3-(k-1)^3}{6^3}.\end{align}$$
A: To compute the probability, we should add the probability of all cases that have a maximum of $m \in \{1,2,3,4,5,6\}$:


*

*Either all three dice $d_1$, $d_2$ and $d_3$ are equal to $m$;

*Or two of them are equal to $m$ and one is strictly less than $m$, there are 3 cases, because either $d_1$, $d_2$ or $d_3$ could be the smallest result;

*Or one of them is equal to $m$ and the two others are strictly less than $m$, there are 3 cases, because either $d_1$, $d_2$ or $d_3$ could be the largest result.
Computing this, results in:
$$P(X = m) = P(d_1 = m) \times P(d_2 = m) \times P(d_3 = m)$$
$$ \quad \quad \quad \quad \quad \quad + 3 \times P(d_1 = m) \times P(d_2 = m) \times P(d_3 < m)$$
$$ \quad \quad \quad \quad \quad \quad + 3 \times P(d_1 = m) \times P(d_2 < m) \times P(d_3 < m) .$$
Hence,
$$ P(X = m) = 1/6^3 + 3 \times (m-1)/6^3 + 3 \times (m-1)^2/6^3$$
$$ = (3m^2 - 3m +1) / 216 .\quad \quad \quad \quad$$
Of course that I considered that the dice rolls are independent, so I multiplied the probability of the outcome of each dice.
