In above problem, we have $6-keys$ out of which $2$ are identical and any of these $2$ can open lock.
One key is dropped. So, the dropped key may be the one that opens lock with a probability of $\frac{2}{6}$ and may be the one that doesn't opens door with probability of $\frac{4}{6}$. (here the problem is divided into $2$ cases)
Case 1: Key that opens door is lost
Now, we are remaining with $5$ keys out of which only one opens the lock.
Case 2: Key that doesn't opens door is lost
Now, we are remaining with $5$ keys out of which $2-keys$ open the lock.
P(lock-opens) = Key that opens door is lost and lock opens (or) Key that doesn't opens door is lost and lock opens
$P(lock-opens) = \frac{2}{6}*\frac{1}{5} + \frac{4}{6}*\frac{2}{5} = \frac{10}{30} = \frac{1}{3}$