The total number of tiles are 8 and there are 3 tiles containing letter I.
The probability for 3 "I" tiles one at a time (without replacement) is,
[tex]\begin{gathered} P(3I^{\prime}s)=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6} \\ =\frac{6}{56\cdot6} \\ =\frac{1}{56} \end{gathered}[/tex]So probability of drawing all three "I" tiles from the bag, one at a time is 1/56.