Finding probability when two dice are rolled

Question

You have two dice. Die one is a standard die with the six faces marked from 1 to 6. The second die has two faces marked with 1, two faces marked with 2 and two faces marked with 3. Both dice are rolled. The probability that the sum of values on the top face of the two dice is greater than 6 is:

A. 6/36

B. 8/36

C. 10/36

D. 12/36

Answer 1

Daniel Mathias already added a good asnwer but I am posting my asnwer since I already began writing it- $$\text{First Die(Left)-SecondDie(Right)}$$ $$6-\{1,1,2,2,3,3\}$$ $$5-\{2,2,3,3\}$$ $$4-\{3,3\}$$

So $$P(Sum>6)=(\frac16\times1)+(\frac16\times\frac46)+(\frac16\times\frac26)=\frac{12}{36}$$

Answer 2

Make a table of possible outcomes: \begin{array}{c|cc} &1&2&3&4&5&6\\ \hline 1&2&3&4&5&6&\boxed7\\ 1&2&3&4&5&6&\boxed7\\ 2&3&4&5&6&\boxed7&\boxed8\\ 2&3&4&5&6&\boxed7&\boxed8\\ 3&4&5&6&\boxed7&\boxed8&\boxed9\\ 3&4&5&6&\boxed7&\boxed8&\boxed9\\ \end{array} Count the number of outcomes with sum greater than $6$. (There are $12$ of them, as marked)

The probability is therefore $\frac{12}{36}$