Assume that two standard dice are being rolled. Let event $\text{A} = \text{{the sum is a multiple of 3}}$ and $\text{event B} = \text{{the sum is a multiple of 4}}$. The $P(A)=\frac{12}{36}$ and the $P(B)=\frac{9}{36}$.  

1. How many outcomes are in the intersection of events A and B?

   Hint (a):

   Make a table or grid with all the sum possibilities for two rolled die,
   i.e. $1+1=2$, $1+2=3$, etc.
   How many outcomes are multiples of both $3$ and $4$?

2. What is P(A or B)?

   Hint (b):

   Use the table or list you created for part (a).

   More Help (b):

   Remember to exclude all outcomes that apply to both events.
   $\text{P(A or B)}=\text{P(A)}+\text{P(B)}-\text{P(A and B)}$

   Answer (b):

   $P(\text{A or B})=\frac{20}{36}$