I'm struggling understanding how conditional probability for a continuously distributed random variable is to be calculated. The task is as follows:
$f(t) = 1/8 * (4-t)$ for $0 < t <= 4 $ and $0$ elsewhere
We know that in this interval exactly one event $E$ will happen, with probability at a specified time $t$ given by the distribution. Find the probability that $E$ will happen in the interval $1<t<=1.5$ given that it has not happened in the interval $0<t<=1$.
I tried setting this up using Bayes' theorem, but it did not seem to be a good solution. How to solve this one?
Try to write it out to make it more clear:
$$ Pr(1<t \leq 1.5| 1<t\leq 4)=\frac{Pr(1<t \leq 1.5, 1<t\leq 4)}{Pr(1<t\leq 4)}=\frac{Pr(1<t \leq 1.5)}{Pr(1<t\leq 4)} $$ where the last equality holds because $(1, 1.5]$ is a subset of $(1, 4]$. Then you can simply calculate the numerator and the denominator using standard ways.