I'm stuck on a problem and I think I'm conceptualising something wrongly.
Jones is waiting for one of two telephone booths occupied by A and B. Let the duration of their calls be an exponential random variable with $\lambda=1/8$. What is the probability that of the three of them, Jones is not the last to finish their call?
My attempt:
Let $T_{1},\;T_2,\;T_3$ be the duration of each of Jones, A and Bs calls respectively. Then these three random variables follow an exponential distribution with $\lambda=1/8$. Assuming that A and B start their calls at exactly the same moment, then the probability we are interested in is: $$\mathbb{P}\:(T_2<T_3,\:\:T_1<T_3-T_2)\:+\mathbb{P}\:(T_3<T_2,\:\:T_1<T_2-T_3)$$ i.e. that A finishes before B, then Jones finished before B, or, B finishes before A, and Jones finishes before A. Since $T_1,\:T_2,\:T_3$ all follow the same distribution you can just take $$2\mathbb{P}\:(T_2<T_3,\:\:T_1<T_3-T_2)$$ for convenience. My problem is that when I try to evaluate that using the cdf for an exponential random variable, I get an expression in terms of $T_2$, and $T_3$ when the answer is : $\frac{1}{2}$ "use the memoryless property". I think I'm missing something here..
First Jones has to wait until one of A and B is finished, then Jones starts his/her call. It doesn't matter which of A and B is finished first, let's say it's A. Now the time until Jones finishes has an exponential distribution with parameter $\lambda$. By the "memoryless" property of the exponential distribution, the time remaining until B finishes also has an exponential distribution with the same parameter $\lambda$, and it is independent of the time until Jones finishes. Thus Jones and B each have probability $1/2$ of being the last one finished.