How can I compute the probability that the component fails between time t and t+δ seconds given it has survived until t? In this question I sought to find the probability of a component failing in a given interval $[t, t + \delta]$ assuming an exponential distribution ($T$ being the time of failure).
\begin{align*}
F(t)= P(T \leq t) = 1 - e^{-\lambda t}
\end{align*}
So, initially I sought $P(t \leq T \leq t + \delta)$. However, after computing some values empirically, I found that the probability of the component failing was decreasing over time which seemed counter-intuitive to me, as with an exponential distribution one would expect that the longer the component lives, the higher probability that it will experience a failure.
My intuition is that the probability I was initially computing was actually the probability that the component would fail in that interval as computed from $t=0$, and the fact that the component has survived until some time $t>0$ must be taken into account using a conditional probability along the lines of: $$P(t \leq T \leq t + \delta ~|~ t \leq T)$$
How can I compute this conditional probability?
 A: 
I found that the probability of the component failing was decreasing over time ...
My intuition is that the probability I was initially computing was actually the probability that the component would fail in that interval as computed from $t=0$

Yes, exactly. The computation of $\int_t^{t+\delta} p(t)\, dt = F(t+\delta)-F(t)$ is the probability that a brand new component is going to fail during the interval $[t,t+\delta]$. And this decreases with larger $t$ precisely because the component is less likely to survive up to time $t$ in the first place.
Bayes' Theorem says $$P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$$
If $A$ is the event $t \leq T \leq t+\delta$ and $B$ is the event $t \leq T$, then $P(B\mid A) = 1$: if $A$ is true then $B$ is necessarily true. We know $P(A) = F(t+\delta)-F(t)$ as explained in your previous question. And $P(B) = 1 - F(t)$ since $F(t)$ is the probability $T \leq t$, $P(B)$ is the probability $T \geq t$, and the probability the real number $T$ is exactly any particular $t$ is zero. So
$$ P(t \leq T \leq t + \delta \mid t \leq T) = \frac{F(t+\delta)-F(t)}{1-F(t)} = \frac{e^{-\lambda t} - e^{-\lambda (t+\delta)}}{e^{-\lambda t}} = 1 - e^{-\lambda \delta} $$
That's right, this probability doesn't depend on $t$ at all. The exponential distribution models a case where an old component is just as likely to fail in any given time interval as a brand new component. We see this a lot in particle physics: particles of the same type act just the same, and how long that particle has existed isn't relevant. A carbon-14 nucleus in a dinosaur fossil and a carbon-14 nucleus formed today in the upper atmosphere have the same chance of decaying in the next year.
A: $$\begin{align}
P(t \leq T \leq t + \delta | t \leq T)&=\frac{P(t \leq T \leq t + \delta \cap t \leq T)}{P(t\le T)}\\
\\
&=\frac{P(t \leq T \leq t + \delta )}{P(t\le T)}\\
\\
&=\frac{F(t+\delta)-F(t)}{1-F(t)}
\end{align}$$
