currently I'm stuck with some probability task (I will provide my solution below): We have interval of 10 minutes (in other words [0; 10]). Two people ("A" and "B") come after each other in random time of interval. By assuming that "A" came before "B" and the distribution is uniform, find the probability that B came in the last 5 minutes of interval (meaning [5; 10])
My solution (I guess it's incorrect, as I end up with probability greater than 1): Let's consider two cases. In the first case, "A" came in the last 5 minutes (with probability 1/2, as it has 5 minutes out of 10), then "B" must came in the last 5 minutes too (with probability 1). So, it's 1/2 * 1 In the second case, "A" came in the first 5 minutes (let's assume, "A" came at the time 'x'). The probability of this situation is 1/2 as well. Then "B" can came in every time of interval [x; 10]. To came in the last 5 minutes, "B" has probability 5/(10 - x). As I understand, we need to take integral with x from interval [0; 5]. So, the whole formula is 1/2 *1 + 1/2 * (integral of (5/(10 - x)), where x from [0; 5]) End up with the answer something like >1. Any ideas of where I've made the mistake, please?
The arrival time for $B$ is not conditionally uniformly distributed when given that for $A$ .
The times for $A$ and $B$ cannot be independent and strictly ordered.
When taking $A,B$ as independent and identically distributed (uniformly over $[0;10]$), you must treat this "assumption" of order as a condition, thus seek to find $\mathsf P(5\leq B\leq 10\mid A<B)$ .
The integration is set up as:
$$\begin{align}\mathsf P(5\leqslant B\leqslant 10\mid A\lt B) ~&=~ \dfrac{~\mathsf P(A<B, 5\leqslant B\leqslant 10)~}{\mathsf P(A\lt B)}\\[2ex]&=~\dfrac{\quad\int\limits_5^{10}\int\limits_0^b f(a)\,f(b)\,\mathrm d a\,\mathrm d b\quad}{\int\limits_0^{10}\int\limits_0^b f(a)\,f(b)\,\mathrm d a\,\mathrm d b}\end{align}$$
$~$
PS: You can save some effort, since by symmetry $\mathsf P(A<B)=1/2$.