Two fair dice are thrown together, and the scores added. What is the probability that the total score is 9, or the individual scores differ by 1, or both?

$\displaystyle P(total score is 9)=\frac{4}{36}=\frac{1}{9}$

$\displaystyle P(individual scores differ by 1)=P{(1,2), (2,3), (2,1), (3,4), (3,2), (4,5),

(4,3), (5,6), (5,4), (6,5)}=\frac{10}{36}=\frac{5}{18}$

$\displaystyle P(both scoring 9 and score differ by 1)=P{(4,5),(5,4)}=\frac{2}{36}=\frac{1}{18}$

Should i treat the question as three seperate parts? The model answer has only 1 answer of 1/3 and that doesn't coincide with any of my answers...