Probabilities of rolling a fair die twice?

Question

a) One of the dice is a 4 b) Sum of the dice equals 4 c) Neither dice is a 4

I get really confused with these exercises (I don't know why)

Answer 1

If you want to roll a die twice, that is the same thing as rolling two dice once. Think of the problem as such.

If we want to solve part A, the easiest thing to do would be to use complementary counting. We can represent the probability that neither of the dice is four as $(5/6)(5/6)$ and the probability that both the dice are four as $(1/6)(1/6)$. By adding these two up, we get $26/36$. Therefore, there is a $(10/36)$ chance we will get exactly one four. If we include the $4$ and $4$, there is an $(11/36)$.

For the sum of dice equals four, we can just use common reasoning and counting. The only way this can happen is with a $(1,3)$, a $(2, 2)$ or a $(3,1)$. The $(1,3)$ and the $(3,1)$ are two different cases, as the dice which rolls each number is switched in the other case. This is three ways, so there is a $3/36$ or $1/12$ probability you will get a sum of 4.

For neither dice to be four, you would simply have $(5/6)(5/6)$, or $P($not a $4)$ * $P($not a $4)$ , or $25/36$.