Since Kelly chooses one tile, places it back in the box, and then choose another tile, we have that the total outcomes are 4x4 = 16.
Now, we can write easily all the outcomes :
[tex]\begin{gathered} \Omega=\mleft\lbrace(1,1\mright),(4,1),(5,1),(8,1), \\ (1,4),(4,4),(5,4),(8,4), \\ (1,5),(4,5),(5,5),(8,5), \\ (1,8),(4,8),(5,8),(8,8)\} \end{gathered}[/tex]notice that on the first row, only the event (8,1) gives us that the sum is greater than 7.
On the second row,we have the same for events (4,4),(5,4) and (8,4)
On the third row only (1,5) gives us a sum lesser than 7, then here we have 3 more events that fit our criteria.
On the last row, all events gives us a sum greater than 7,then we have another 4 events.
Then, the probability that the sum of two chosen tiles is greater than 7 is:
[tex]\frac{1}{16}+\frac{3}{16}+\frac{3}{16}+\frac{4}{16}=\frac{11}{16}[/tex]