clarify basic probability question

Question

You have a loaded die.

$A =$ even numbers

$B =$ odd numbers

$P(A)=3/4$

$P(B)=1/4$

$P(\{1\})=1/12$

$P(\{2\})=1/4$

How was $P(\{1\})$ and $P(\{2\})$ calculated? Is it beacuse there are 3 odds and 3 evens so the chance of 1 odd (1) is 1/3(1/4)=1/12 and likewise, the chance of an even (2) is 1/3(3/4)=1/4?

Answer 1

$P(\{1\}) = P(B\cap \{1\}) = P(B)\cdot P(\{1\}|B) = \dfrac{1}{4}\cdot \dfrac{1}{3} = \dfrac{1}{12}$, and similarly the other one is: $\dfrac{3}{4}\cdot \dfrac{1}{3} = \dfrac{1}{4}$

Answer 2

You are right.

If you roll the dice 12 times, you'll get 9 times 2, 4 or 6, and 3 times 1, 3 or 5. Said differently, three 2, three 4, three 6, one 1, one 3 and one 5.

On average, of course, and assuming that the even and odd numbers are equiprobable inside their category.