Are we sure that this is correct? The way I see it the possible outcomes of rolling die 1 (the special die) are:{1,2,3} with each having a \(\displaystyle \frac{1}{3}\) probability. And for die 2 (the regular, six-side die) our possible outcomes are:{1,2,3,4,5,6} with each number having a \(\displaystyle \frac{1}{6}\)

So for part A:

(We'll assume that in each of these pairs the first number listed is from the special die and the second number from the regular die)

The possible outcomes for achieving a sum of 4 are:{(1,3),(2,2),(3,1)}.

Okay, so this is equivalent to \(\displaystyle pr(1\cap 3)+pr(2\cap 2)+pr(3\cap 1)\)

Lastly, for C we're looking for the probability that we will roll the same number on both dies

\(\displaystyle pr(1\cap 3)= pr(1)*pr(3)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(2\cap 2)= pr(2)*pr(2)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(3\cap 1)= pr(3)*pr(1)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

So then:

\(\displaystyle pr(1\cap 3)+pr(2\cap 2)+pr(3\cap 1)= \frac{1}{18}+\frac{1}{18}+\frac{1}{18}= \frac{3}{18}=\frac{1}{6}\)

For B we're looking for the probability that neither die is a 3.

So for the special die, the probability that you won't roll a 3 is \(\displaystyle pr(3')=\frac{2}{3}\) and for the regular die, the probability that you won't roll a 3 is \(\displaystyle pr(3')=\frac{5}{6}\). So now that probability that you won't roll a 3 on either die is

\(\displaystyle pr(3'\cap 3')=\frac{2}{3}*\frac{5}{6}=\frac{10}{18}=\frac{5}{9}\)

Lastly, for part C we're looking for the probability that we'll roll the same number on both dice meaning we'll get {(1,1),(2,2),(3,3)}.

\(\displaystyle pr(1\cap 1)= pr(1)*pr(1)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(2\cap 2)= pr(2)*pr(2)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

\(\displaystyle pr(3\cap 3)= pr(3)*pr(3)=\frac{1}{3}*\frac{1}{6}=\frac{1}{18}\)

So then:

\(\displaystyle pr(1\cap 1)+pr(2\cap 2)+pr(3\cap 3)= \frac{1}{18}+\frac{1}{18}+\frac{1}{18}= \frac{3}{18}=\frac{1}{6}\)

This seems correct to me but of course I could be wrong. If I am, I'd love for someone to steer me in the right direction