123. On the highway, there is an automatic traffic light that shows green light for 1 minute for vehicles and red light for 45 seconds, then again 1 minute of green light and 45 seconds of red light, and so on. A car passes along the highway at a random moment in time, unrelated to the operation of the traffic light. Find the probability that the car will pass by the traffic light without stopping.
S o l u t i o n. Let's consider the random variable $T$ - the moment of time when a car passes by the traffic light within an interval equal to the period of the traffic light color changes. The period of the traffic light color changes is $(1+3 / 4)=7 / 4$ minutes. The random variable $T$ is uniformly distributed on the interval $[0 ; 7 / 4]$. All possible values $t$ of this variable belong to the interval with the lower boundary $a=0$ and the upper boundary $b=7 / 4$. On this interval, the probability density function of the variable $T$ is

$$
f(t)=\frac{1}{b-a}=\frac{4}{7}
$$

Outside this interval, $f(t)=0$. The graph of the function $f(t)$ is shown in Fig. 21.

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-151.jpg?height=286&width=385&top_left_y=1451&top_left_x=522)

Fig. 21

The probability of the random variable $T$ falling within the interval $[0; 7 / 4]$ is equal to the area of the rectangle $\mathrm{OABC}$, which is equal to one.

For the car to pass by the traffic light without stopping, it is sufficient for the moment of time when the car passes on the highway to fall within the interval during which the green light is on, i.e., the interval $(0 ; 1)$. The probability of the random variable $T$ falling within this interval is equal to the area of the rectangle OADE, which is equal to 4/7. Thus, the probability of the car passing by the traffic light without stopping is $P(00)
$$

where $a$ is the expected value, $\sigma$ is the standard deviation of $X$.

The probability that $X$ will take a value within the interval $(\alpha ; \beta)$ is

$$
P(\alpha<X<\beta)=\Phi\left(\frac{\beta-a}{\sigma}\right)-\Phi\left(\frac{\alpha-a}{\sigma}\right)
$$

where

$$
\Phi(x)=\frac{1}{\sqrt{2 \pi}} \int_{0}^{x} e^{-x^{2} / 2} d x-\text { the Laplace function. }
$$

The probability that the absolute value of the deviation of a normally distributed random variable $X$ from its expected value $a$ does not exceed a positive number $\varepsilon$ is

$$
P(|X-a| \leq \varepsilon)=2 \Phi\left(\frac{\varepsilon}{\sigma}\right)
$$