Solution: Treat the two letters as a single entity. This gives $4!$ arrangements for the entities (the letter pair and 3 manuscripts). The letters within the pair can be ordered in $2!$ ways. Total favorable arrangements: $4! imes 2!$. Total possible arrangements: $5!$. The probability is $rac4! imes 2!5! = rac25$. $oxed{\dfrac25}$ - ToelettAPP