# Traffic lights probability

Ive been asked the following :

Two consecutive traffic lights have been synchronized to make a run of green lights more likely. In particular, if a driver finds the first light to be red, the second light will be green with probability 0.9, and if the first light is green the second will be green with probability 0.71 . If the probability of finding the first light green is 0.62 , find the probability that a driver will find both lights green.

Here is how I've modeled the probabilities :

Defining :

a : first light red

b : second light green

c : first light green

Therefore:

$$P(a\cap b)= 0.9$$ $$P(c\cap b)= 0.71$$

If $$P(c)= 0.62$$ what is $$P(c\cap b)$$ ?

I calculate the probability to be $$P(c\cap b)P(c)=(.71)(.62)=.4402$$ I doubt this is correct but not sure what other path to take ?

Referencing this question which is similar : Traffic Light Probability

• I think your notation is not correct: I think $P(c\cap b)$ would already mean: Probability of both lights being green. The calculation itself however is correct. Commented Nov 11, 2018 at 9:38
• You should write $P(b | a)= 0.9$, $P(b | c)= 0.71$, so that $P(c \cap b) = P(b | c)P(c)=(.71)(.62)=.4402$
– N74
Commented Nov 11, 2018 at 9:50
• @N74 I calculated correct probability value (.4402) using incorrect logic ? Commented Nov 11, 2018 at 9:52
• And I'd suggest you to use better names for the events: let's say that the name $g$ is for green and $r$ is for red and we use a subscript for the semaphore, you can express your probabilities as $P(g_2 | r_1)= 0.9$, $P(g_2 | g_1)= 0.71$, so that $P(g_2 \cap g_1) = P(g_2 | g_1)P(g_1)=(.71)(.62)=.4402$ that is easier to read.
– N74
Commented Nov 11, 2018 at 9:58

You are asked to find the probability that the first light and the second light are green.

For two events $$A$$ and $$B$$ we have

$$\mathsf P(A\cap B)=\mathsf P(A\mid B)\mathsf P(B)$$

where $$\mathsf P(A\cap B)$$ denotes the probability that $$A$$ and $$B$$ both occur and $$\mathsf P(A\mid B)$$ denotes the probability that $$A$$ happens, given that $$B$$ happens.

Using the events you have already defined we have

$$\mathsf P(B\cap C)=\mathsf P(B\mid C)\mathsf P(C)=0.71\cdot0.62=0.4402$$

Note in probability we usually capitalize letters which represent random variables.