I have the following scenario provided about the rolling of two dice.
The question I'm trying to answer states that:
Calculate the probability of ($X_1$ − 1)+ > $X_2$
I'm not too sure about the proper way of doing it, but I attempted something like:
$X_2$ = (1,2,3,4,5,6) representing 6 sides.
$X_1$ - 1 = (0,1,2,3,4,5) for 6 sides too but since we are subtracting by 1, we have the following valid sides.
Therefore, the probability is 3/6 = 1/2 since sides 3,4,5 in $X_1$-1 is the only sides that is greater than some sides in $X_2$.
I'm not too sure if I'm doing it correctly, hence I would appreciate some help on this.
We have to consider numbers in pairs, that is count how many pairs of $(x,y)$ satisfies $(x-1)_+ > x_2$ then divide it by $36$.
We can also use the law of total probability.
\begin{align} P(\lfloor X_1-1 \rfloor_+ > X_2) &= P(X_1-1 > X_2)\\ &= \sum_{x_2=1}^6 P(X_1>X_2+1|X_2=x_2)P(X_2=x_2)\\ &=\frac16 \sum_{x_2=1}^6P(X_1 > x_2+1)\\ &=\frac16 \cdot \sum_{x_2=1}^4P(X_1>x_2+1)\\ &=\frac16 \sum_{x_2=1}^4 \frac{5-x_2}{6}\\ &= \frac16 \cdot \frac{1+2+3+4}{6}\\ &= \frac{10}{36} \end{align}