41. A signal detector receives signals from two devices, and the arrival of each signal is equally likely at any moment within a time interval of duration $T$. The moments of signal arrival are independent of each other. The detector triggers if the difference between the moments of signal arrival is less than $t(t<T)$. Find the probability that the detector triggers within the time $T$, if each device sends one signal.
Solution. Let the moments of signal arrivals from the first and second devices be denoted by $x$ and $y$, respectively. According to the problem's condition, the following double inequalities must be satisfied: $0 \leq x$ and $x - t \leq y \leq x$, or equivalently,

$$
\begin{array}{lll}
y \leq x \\
y > x - t & \text { when } & y < x .
\end{array}
$$

Inequality (*) is satisfied for the coordinates of points in the figure $G$ that lie above the line $y = x$ and below the line $y = x + t$; inequality (**) holds for points located below the line $y = x$ and above the line $y = x - t$.

As can be seen from Fig. 4, all points whose coordinates satisfy inequalities (*) and (**) belong to the shaded hexagon. Thus, this hexagon can be considered as the figure $g$, the coordinates of whose points are the favorable moments of time $\boldsymbol{x}$ and $y$ for the alarm to trigger.

The desired probability is

$$
P=\frac{\Pi \pi \cdot g}{\Pi \pi: G}=\frac{T^{2}-2(T-t)^{2} \cdot 2}{T^{2}}=\frac{t(2 T-t)}{T^{2}}
$$