Can't we look it in the below fashion:
With dividing the code in two blocks say 90% and 10 % of code,
total reliability = success of 1st block + success of 2nd block
= 0.9*0.9 + (any success %)*0.1
in any case the total reliability will be atleast 0.81?
Just trying to approach the solution in different way, please comment if I am doing anything wrong.