A four sided die (tetrahedron) and a $6$ sided die (cube) are rolled together.

Question

A four sided die (tetrahedron) and a $6$ sided die (cube) are rolled together. The values on the bottom of each are observed.

What is the probability of a $1$ or $3$ being rolled, but not both?

Answer 1

Hint:

The total number of outcomes $(n)=6\times4=24$

Now those possible outcomes are $$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\}$$

Now from the above possible outcomes try to figure out the probability of $1$ or $3$ being rolled.

Answer 2

You seem to be on the right track for the most part, but you might need to read things a bit more closely.

Based on the exact semantics of the question you posed, the only permutations excluded by the "but not both" restriction are $(1, 3)$ and $(3, 1)$. The event of "a $1$ or a $3$, but not both" includes $(1, 1)$ and $(3, 3)$ because in both those cases, you rolled either a $1$ or a $3$, but you did not roll both a $1$ and a $3$. The event doesn't exclude cases where a $1$ or a $3$ was rolled on both dice; it excludes cases where both a $1$ and a $3$ were rolled on the dice.

The cases $(3, 1)$ and $(1, 3)$ do not satisfy the event because both a $1$ and a $3$ were rolled on the dice. The case $(1, 1)$ does satisfy the event because a $1$, but not a $3$, was rolled; and the case $(3, 1)$ does satisfy the event because a $3$, but not a $1$, was rolled. If the event was, "a $1$ or a $3$ was rolled on one, but not both of the dice", you would be correct, but the event is, "a $1$ or a $3$, but not both [a $1$ and a $3$], was rolled on the dice".