Rafael spins the pointers of the two spinners shown at the right. Find the probability of each possible sum.

Answer:
P(sum 2)=1/6
P(sum 3)=1/3
P(sum 4)=1/3
P(sum 5)=1/6
Step-by-step explanation:
[tex]\left|\begin{array}{c|ccc}&1&2&3\\---&---&---&----\\1&2&3&4\\2&3&4&5\end{array}\right|[/tex]
On the table, there are a a total of 6 outcomes.
Therefore:
P(sum 2)=1/6
P(sum 3)=2/6=1/3
P(sum 4)=2/6=1/3
P(sum 5)=1/6