In case you don't know the general form of computing such an expectation, follow THIS link.
Thanks for your reply captain black. I thought that P(T/T<=t) will become
P(T) because the fact that the waiting time was <t has no effect on the distribution of the time you will wait due to the memorylessness property, but you have P(T/T<=t)=P(T,T<=t)/P(T<=t) written as P(T)/P(T<t).
I know you are most likely correct, but i would like to understand the logic as i am not getting how you arrived at this.