According to the description the outcome space is a set of pairs \(\displaystyle (x,y)\) where \(\displaystyle 1\le x\le y~\&~1\le y\le 8\).
Then we get \(\displaystyle \text{result }=xy\text{ if }x\text{ is odd}\) OR \(\displaystyle \text{result }=x+y\text{ if }x\text{ is even}\).
Now if the result is \(\displaystyle 6\) then the pairs could be \(\displaystyle (1,6),~(2,4),~(3,2),~\text{ or }~(4,2)\)
But the probability of any pair is \(\displaystyle \frac{1}{32}\)