I need some help with a specific probability question. It's about radioactive decay. The probability that a radioactive particle will decay in the interval $0\leq t\leq T$ is
$$
P( [0,T] )=\frac{PART}{WHOLE}= \frac{1}{W} \int_0^T e^{-\ln(2)t} dt
$$
Where $W$ is $\int_0^\infty e^{-\ln(2)t} dt$ (which I evaluated to be $\frac{1}{\ln2}$).
Q.: What is the probability that it survives to time $t=1$, but decays some time during the interval $1\leq t\leq 2$? (Only give the integral formula and approximate it with a calculator.)

I gave this problem a go and reasoned that such probability would be given by:
$$
P([1,2]) = \frac{1}{W} \int_1^2 2^{-t} dt = \frac{1}{4}
$$
But this is clearly wrong, since one can evaluate such integral pretty easily, without the need for a calculator. I've reviwed the lessons on probability with little to no improvements. Could anyone help me?
 A: Since the particle can only decay once, the probability of decay in a particular interval is equal to the expected number of decays in that interval.  Because of this, you can use linearity of expectation to write the answer exactly as you did:
$$
P([1,2])=P([0,2])-P([0,1])=\frac{\int_{1}^{2}e^{-t \log 2} dt}{\int_{0}^{\infty}e^{-t \log 2} dt}=\frac{-e^{-t\log 2}\vert_{1}^{2}}{-e^{-t\log 2}\vert_{0}^{\infty}}=\frac{2^{-1}-2^{-2}}{2^{-0}-2^{-\infty}}=\frac{1}{4}.
$$
You may alternately want to use the fact that decay is a memory-less process, so the probability of decaying in any interval $[a,b]$ (conditioned on survival up to time $a$) is just a function of $b-a$.  So, since
$$
P([0,1])=\frac{1}{2},
$$
you also have $P(t\in[1,2] \;|\;t\ge 1)=1/2$, and
$$
P([1,2])=P(t\in [1,2] \;|\; t\ge 1)\cdot P(t\ge 1)=P(t\in [1,2] \;|\; t\ge 1)\cdot \left(1 - P([0,1]\right)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.
$$
(But note that this approach only works if the decay rate is constant over time.  Your initial approach, writing the result as an integral over the specific interval $[1,2]$, is more generally correct.)
