Two people, A and B, agree to meet at a location sometime within the interval $[0, T]$. The first person to arrive waits for the other person, but if the other person does not arrive within time $t$ (where $t < T$), the first person leaves. The arrival times of A and B are uniformly random and independent within the interval $[0, T]$. The problem asks for the probability that A and B successfully meet.

1. Problem Description

Two people, A and B, agree to meet at a location sometime within the interval $[0, T]$. The first person to arrive waits for the other person, but if the other person does not arrive within time $t$ (where $t < T$), the first person leaves. The arrival times of A and B are uniformly random and independent within the interval $[0, T]$. The problem asks for the probability that A and B successfully meet.

2. Solution Steps

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

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

For A and B to meet successfully, the difference between their arrival times must be less than or equal to $t$, i.e., $|X - Y| \le t$. This condition can be rewritten as $-t \le X - Y \le t$, which means $X - t \le Y \le X + t$.

We need to find the probability $P(|X - Y| \le t)$. This can be represented as a double integral over the region where the condition is satisfied. The region of interest is the square $[0, T] \times [0, T]$ in the $xy$-plane, and we want to find the area where $x - t \le y \le x + t$. We compute the probability by integrating the joint density function over this region:

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

The area of the region where $|x - y| \le t$ within the square $[0, T] \times [0, T]$ can be found by calculating the area of the square and subtracting the area of the two triangles where $|x - y| > t$.

The area of the square is $T^2$. The two triangles are defined by $y > x + t$ and $y < x - t$.

The triangle where $y > x + t$ has vertices $(0, t), (0, T), (T-t, T)$. The area of this triangle is $\frac{1}{2}(T-t)^2$.

The triangle where $y < x - t$ has vertices $(t, 0), (T, 0), (T, T-t)$. The area of this triangle is $\frac{1}{2}(T-t)^2$.

So, the area of the region where $|x - y| \le t$ is $T^2 - (T-t)^2$.

Therefore, the probability is:

$P(|X - Y| \le t) = \frac{1}{T^2} [T^2 - (T-t)^2] = \frac{T^2 - (T^2 - 2Tt + t^2)}{T^2} = \frac{2Tt - t^2}{T^2} = \frac{2t}{T} - \frac{t^2}{T^2}$

$P(|X - Y| \le t) = 1 - \frac{(T-t)^2}{T^2} = 1 - (1 - \frac{t}{T})^2 = 1 - (1 - 2\frac{t}{T} + \frac{t^2}{T^2}) = \frac{2t}{T} - \frac{t^2}{T^2}$

3. Final Answer

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