Add solution for 3f

exam/ex.tex

@@ -185,7 +185,11 @@
       a preimage is $1/2^{55+90} = 1/2^{145}$. This is due to the fact that the
       first $\unit[55]{bits}$ and the last $\unit[90]{bits}$ have to be zeros.
 
-    \item \TODO
+    \item An adversary does not gain additional information by using inverse
+      queries additionally to forward queries. Forward queries have a better
+      probability of being successful at breaking preimage resistance and thus
+      an adversary should focus on that. The total probability is thus
+      $1/2^{101}$.
 
   \end{enumerate}