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
Best Answer
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.