## 19. Red and Green

On a large one-way street, two traffic lights are located one after another; each traffic light is designed so that the period when the green light is on constitutes two-thirds of the total operating time of the traffic light. A motorist noticed that when he, moving at a normal speed, passes the first traffic light on green, in 3 out of 4 cases the second traffic light does not hold him up. Suppose now that the motorist passes the first traffic light when it is red. What is the probability that the second traffic light will also be red?
19. Let's consider what a sequence of traffic light signals might be. There are four possibilities: "both green"; "the first is green, the second is red"; "the first is red, the second is green"; "both are red". Let $p_{1}, p_{2}, p_{3}$, and $p_{4}$ be the probabilities of the respective four events. Since each traffic light gives a green light $2 / 3$ of the time, we get

$$
p_{1}+p_{2}=2 / 3, \quad p_{1}+p_{3}=2 / 3
$$

On the other hand, if the first traffic light gives a green light, the probability that the second (when the motorist approaches it) will also give a green light is 3/4:

$$
p_{1} /\left(p_{1}+p_{2}\right)=3 / 4
$$

Finally, it is obvious that

$$
p_{1}+p_{2}+p_{3}+p_{4}=1
$$

Solving the system of four equations with four unknowns, we find

$$
p_{1}=1 / 2 ; \quad p_{2}=p_{3}=p_{4}=1 / 6
$$

Thus, the probability of encountering a red signal at the second traffic light given that the first traffic light was also red is

$$
\frac{p_{4}}{p_{3}+p_{4}}=\frac{1}{6}: \frac{2}{6}=\frac{1}{2}
$$