The second pick is dependent on the first one.
Let [tex]a[/tex] be the number of tiles with the letter A, and [tex]n[/tex] be the total number of tiles.
With the first pick, you have probability [tex]\frac{a}{n}[/tex] to pick an A, because you have [tex]a[/tex] "good" cases out of [tex]n[/tex] possible cases.
The second pick depends on the first one: