Suppose $X$ is a exponential random variable with $exp(1)$, i.e., the density function $f(x)=f_X(x)=e^{-x}$, the cumulative distribution function $F(x)=F_X(x)=1-e^{-x}$, where $x>0$. Let $Y=[X+1]$ (the floor function, i.e., $Z-1<[Z]\leq Z$), then find the conditional distribution of $X-5$ given $Y\geq 5$.
I tried to solve it as follows:
Let $S(t)=P(X-5>t|Y\geq 5)$. By here, we get $Y$ is a geometric random variable, with $P(Y=n)=(1-p)^{n-1}p$. Then we can compute $P(Y\geq 5)=(1-p)^4$ and $S(t)=P(X-5>t|Y\geq 5)=\frac{P(X-5>t, Y\geq 5)}{(1-p)^4}$. How about next?
Hint: $Y \geq 5$ iff $X+1 \geq 5$ or $X \geq 4$.