Constant default probability problem (apologies for the question title, I couldn't think what to put)
My question regards this problem proposed at an interview:
"If the probability of seeing a car on the highway in 30 minutes is 0.95, what is the probability of seeing a car on the highway in 10 minutes? (assume a constant default probability)"
Also assume "seeing a car" means seeing at least one car.
Let $P(x)$ denote the probability for $x$ to happen.
One way to tackle this problem:
let $x=$ "seeing a car on the highway in 30 minutes".
let $y=$ "seeing a car on the highway in 10 minutes".
$P(x) = 0.95 \Rightarrow P(\lnot x) = 0.05$
We know that $P(\lnot x) = (P(\lnot y))^3$
Hence $P(y) = 1 - \sqrt[3]{P(\lnot x)}$
This is when my query comes into place:
If you wanted to find out the probability of seeing a car within, say 6.34 minutes, would the result below hold true?
$z = $ "seeing a car on the highway in 6.34 minutes"
$P(z) = 1 - \sqrt[\frac{30}{6.34}]{P(\lnot x)}$
 A: Probably yes. What you are describing is usually modeled as a poisson process. If arrivals follow a poisson process of rate $\lambda$ then the number of arrivals in an interval of size $t$ as a poisson variable of rate $\lambda t$. Now, the (constant) rate $\lambda$ poisson process model applies when the rate of arrivals is a constant $\lambda$, which is how I interpret "(assume a constant default probability)", and when the probability of an arrival in a given interval of time is independent of the probability of arrival in other disjoint intervals of time, which is what you are assuming for the specific 3 intervals of 10 min when you say "We know that $P(\lnot x) = (P(\lnot y))^3$".
So say you are told the probability of seeing a car show up in 30 minutes is $p=.95$. Under the model a car arrival in 30min is a poisson $30\lambda$ random variable, so the probability of seeing 1 or more arrivals is $1-e^{-30\lambda},$ so $\lambda=-\log (1-p)/30$ is the rate of the process. So then in some other interval of length say $t=6.34$ (but $t$ could be greater than 30min also, e.g.) you have the probability of seeing a car is the probability a poisson $\lambda t$ variable exceeds $0$, which is $1-e^{-\lambda t}=1-(1-p)^{t/30}.$
