What is the probability that station A transmit before station B?

Question

Given two stations, station A transmit in range time [0, T1], and station B transmit in range time [0,T2]. And let's assume that the transmission times for station A and station B are uniformly distributed within their respective time ranges.
What is the probability that station A transmit before B while T2 > T1 ?

I saw this solution to this question:
Since the transmission times are uniformly distributed, the probability that A transmits before B is equal to the ratio of the length of time during which A can transmit before B (from 0 to T1) to the length of time during which B can transmit (from 0 to T2).

Mathematically, the probability P(A transmits before B) is given by: P(A transmits before B) = (T1 - 0) / (T2 - 0) = T1 / T2

But I'm not fully understand why it's correct... Why it's equal to the ration between their ranges?

Thank you..

Answer 1

The required probability is given as follows: \begin{align*} \Pr(A \text{ transmits before } B) & = \int_{0}^{\min\{T_1, T_2\}} \underbrace{\left(1 - \frac{t}{T_2}\right)}_{\text{Prob. that } B \text{ transmits after time }t } \cdot \underbrace{f_A(t)}_{\text{PDF of } A} \ \mathrm{d}t \\ & = \int_{0}^{\min\{T_1, T_2\}} \left(1 - \frac{t}{T_2}\right) \cdot \frac{1}{T_1} \ \mathrm{d}t \\ & = \frac{\min\{T_1, T_2\}}{T_1} - \frac{\min\{T_1, T_2\}^2}{2T_1T_2}. \end{align*} This can be simplified as \begin{align*} \Pr(A \text{ transmits before } B) & = \begin{cases} 1 - \dfrac{T_1}{2T_2} & \text{ if } T_1 \leq T_2, \\ \dfrac{T_2}{2T_1} & \text{ if } T_2 < T_1.\end{cases} \end{align*}

Answer 2

You can also use a reasoning with less computation. Call $t_1$ and $t_2$ the transmission times for A and B.

Let's assume $T_1 \leq T_2$.

\begin{align*} \ & Pr(A \text{ transmits before } B) = Pr(t_1<t_2) \\ & = Pr((t_1<t_2) \cap (t_2<T_1)) + Pr((t_1<t_2) \cap (t_2>T_1)) \\ & = Pr((t_1<t_2) | (t_2<T_1))*Pr(t_2<T_1) + Pr((t_1<t_2) | (t_2>T_1))*Pr(t_2>T_1) \\ &= 1/2*T_1/T_2+1*(T_2-T_1)/T_2\\ &= 1 - \frac{T_1}{2T_2} \end{align*}

By symmetry, if $T_1 > T_2$, $\Pr(A \text{ transmits before } B)=\frac{T_2}{2T_1}$