Conditional Probabilities Poisson Process

Question

If I let ${X(t); t>=0}$ be a Poisson process having rate parameter $\lambda = 2$. I'm supposed to determine the probability:

Pr{${X(1)>=2 | X(1) >=1}$}

My solution: I looked at this as satisfying the memoryless property Pr{${X>s+t|X>t}$}=P{${X>s}$}

Thus the answer I get is Pr{${X(1)>1}$}=$e^-2$

But the book gives an answer of $(1-3e^-2) \over (1-e^-2)$

Anyone know where I went wrong/what I am missing?

Answer 1

$$P\left(X\geq2\mid X\geq1\right)=\frac{P\left(X\geq2\wedge X\geq1\right)}{P\left(X\geq1\right)}=\frac{P\left(X\geq2\right)}{P\left(X\geq1\right)}=\frac{1-3e^{-2}}{1-e^{-2}}$$

Memoryless property is dealing with $T_1$ denoting the first time an event occurs: $$P(T_1>s+t\mid T_1>s)=P(T_1>t)$$