how to solve this continuous random variable problem? A student has two calculators – an old and a new one. The time it takes the old calculator to provide an answer to a difficult integration problem is exponentially distributed with mean 8 seconds. The new one gives the answer uniformly between 2 seconds and 6 seconds. If the integration was computed within 4 seconds, what is the probability that the student used the old calculator?
 A: Let $T$ be the time it took to answer the question, $E$ the event "old calculator was used," and $f_T$, $f_{T\mid E}$, $f_{T\mid E^c}$ the probability densitity of $T$, the conditional density of $T$ given $E$, and the conditional density of $T$ given $E^c$. Assume the prior probability that the old calculator was used is $p:=P(E)\in(0,1)$. Using Bayes' rule, we have
\begin{align}
\mathbb P(E\mid T\leqslant 4) &= \frac{\mathbb P(T\leqslant 4\mid E)\mathbb P(E)}{\mathbb P(T\leqslant 4\mid E)\mathbb P(E) + \mathbb P(T\leqslant 4\mid E^c)\mathbb P(E^c)}\\
&= \frac{\mathbb P(T\leqslant 4\mid E)p}{\mathbb P(T\leqslant 4\mid E)p + \mathbb P(T\leqslant 4\mid E^c)(1-p)}.
\end{align}
Now,
\begin{align}
\mathbb P(T\leqslant 4\mid E) &= \int_{(-\infty,4]} f_{T\mid E}\\
&= \int_0^4 \frac18 e^{-\frac18 t}\,\mathsf dt\\
&= 1-e^{-\frac12}
\end{align}
and
\begin{align}
\mathbb P(T\leqslant 4\mid E^c) &= \int_{(-\infty,4]}f_{T\mid E^c}\\
&= \int_2^4 \frac14\,\mathsf dt\\
&=\frac12,
\end{align}
so it follows that
\begin{align}
\mathbb P(E\mid T\leqslant 4) &= \frac{p\left(1-e^{-\frac12}\right)}{p\left(1-e^{-\frac12}\right) + \frac12(1-p)}.
\end{align}
