Person A and Person B agree to meet at a location within the time interval $[0, T]$. The first person who arrives waits for time $t$ before leaving, where $t < T$. Given that each person's arrival time is uniformly random and independent, what is the probability that they successfully meet?

1. Problem Description

Person A and Person B agree to meet at a location within the time interval $[0, T]$. The first person who arrives waits for time $t$ before leaving, where $t < T$. Given that each person's arrival time is uniformly random and independent, what is the probability that they successfully meet?

2. Solution Steps

Let $X$ be the arrival time of person A and $Y$ be the arrival time of person B. Both $X$ and $Y$ are uniformly distributed in the interval $[0, T]$. Thus, the joint probability density function is given by

$f(x, y) = \frac{1}{T^2}$, for $0 \le x \le T$ and $0 \le y \le T$.

They will meet if the absolute difference between their arrival times is no more than $t$, i.e., $|X - Y| \le t$. This inequality can be written as $-t \le X - Y \le t$, or $X - t \le Y \le X + t$.

We need to find the probability that $X - t \le Y \le X + t$. This is given by the integral of the joint probability density function over the region where this inequality holds.

$P(|X - Y| \le t) = \iint_{|x-y| \le t} f(x, y) dx dy = \frac{1}{T^2} \iint_{|x-y| \le t} dx dy$

The region of integration is defined by $0 \le x \le T$, $0 \le y \le T$, and $x - t \le y \le x + t$.

We can compute the area of this region by subtracting the area of the two triangles where they don't meet from the total area $T^2$.

The region where they don't meet is defined by $y > x + t$ or $y < x - t$.

The area of the triangle where $y > x + t$ is $\frac{1}{2}(T-t)^2$.

The area of the triangle where $y < x - t$ is $\frac{1}{2}(T-t)^2$.

So, the area where they do not meet is $(T-t)^2$.

The area where they do meet is $T^2 - (T-t)^2 = T^2 - (T^2 - 2Tt + t^2) = 2Tt - t^2$.

Then the probability that they meet is

$P(|X - Y| \le t) = \frac{2Tt - t^2}{T^2} = \frac{t(2T - t)}{T^2} = 2\frac{t}{T} - \frac{t^2}{T^2}$.

3. Final Answer

The probability that A and B successfully meet is $\frac{2Tt - t^2}{T^2} = 2\frac{t}{T} - \frac{t^2}{T^2}$.