3 urns, each with 4 balls. select one ball from each Three urns are labeled $1,2,3$. Each urn contains $4$ balls labeled $1,2,3,4$. A ball is drawn from each urn such that any ball is equally likely to be drawn. The number on the ball is compared to the number of the urn.
Let $X$ be the number of balls with label higher than the urn it was drawn from. For example if balls $3,3,1$ where drawn from urns $1,2,3$ respectively, $X$ would have a value of $2$.
Calculate the discrete probability mass function for $X$.
 A: Examine each urn individually, and for urn $i$ for $i\in\{1,2,3\}$ let $X_i=1$ if the the number of the drawn ball is greater than the urn label, or else $X_i=0$. 
For urn $1$, the we have $P(X_1=0)=\frac{1}{4}$ and $P(X_1=1)=\frac{3}{4}$ 
For urn $2$, the we have $P(X_2=0)=\frac{1}{2}$ and $P(X_2=1)=\frac{1}{2}$ 
For urn $3$, the we have $P(X_3=0)=\frac{3}{4}$ and $P(X_3=1)=\frac{1}{4}$ 
The task is to calculate the probability mass function of $X_1+X_2+X_3$. 
As we have only three urns, we can exhaustively list all eight possible outcomes:
$$P(X_1=0, X_2=0,X_3=0)=\frac{1}{4}\frac{1}{2}\frac{3}{4}=\frac{3}{32}\\
P(X_1=0, X_2=0,X_3=1)=\frac{1}{4}\frac{1}{2}\frac{1}{4}=\frac{1}{32}
\\
P(X_1=0, X_2=1,X_3=0)=\frac{1}{4}\frac{1}{2}\frac{3}{4}=\frac{3}{32}
\\
P(X_1=0, X_2=1,X_3=1)=\frac{1}{4}\frac{1}{2}\frac{1}{4}=\frac{1}{32}
\\
P(X_1=1, X_2=0,X_3=0)=\frac{3}{4}\frac{1}{2}\frac{3}{4}=\frac{9}{32}
\\
P(X_1=1, X_2=0,X_3=1)=\frac{3}{4}\frac{1}{2}\frac{1}{4}=\frac{3}{32}
\\
P(X_1=1, X_2=1,X_3=0)=\frac{3}{4}\frac{1}{2}\frac{3}{4}=\frac{9}{32}
\\
P(X_1=1, X_2=1,X_3=1)=\frac{3}{4}\frac{1}{2}\frac{1}{4}=\frac{3}{32}$$
Thus, we have
$$P(X_1+X_2+X_3=0)=\frac{3}{32}\\
P(X_1+X_2+X_3=1)=\frac{13}{32}
\\
P(X_1+X_2+X_3=2)=\frac{13}{32}
\\
P(X_1+X_2+X_3=3)=\frac{3}{32}$$
