Question: We throw a die twice. Let $X$ be the sum of the results. Let $B = \{ 2, 5 \}$. What is $X^{-1}(B)$?
Answer sheet says: Since the result is an integer number, we look for values $3$ and $4$. If we take $\Omega = \{ 1, 2, ..., 6 \}^{2}$, then $X^{-1}(B) = \{ \{1,2\}, \{2,1\}, \{2,2\}, \{1,3\}, \{3,1\} \}$.
I have to admit, I'm confused. Surely the inverse image should be the possible numbers that result in the sum $2$ or $5$? In other words, $X^{-1}(B) = \{ \{1,1\}, \{1,4\}, \{4,1\}, \{2,3\}, \{3,2\} \}$. Can someone explain why this is not the case? Why do we care about "values 3 and 4"?