Accidents occur in a factory at a rate of 2 per week: $(X \sim Po(2)).$
What is the probability that the time to the first accident is greater than 2 weeks?
What is the probability that the time to the first accident is less than 2 days?
What I've tried so far:
There are 4 accidents every 2 weeks, so $X \sim Po(4)$. I think the question is meaning to ask what the probability is that there will be $0$ accidents in the first 2 weeks, so the poisson rv would be $0$, making the equation --
$$P(X = 0) = \frac{e^{-4}4^0}{0!} = 0.0183$$
for the second one, there are $0.5714$ accidents in $2/7$ of a week (2 days), so $X \sim Po(0.5174)$, but I think the question is asking what the probability of one accident on the first day will be, so that would make $0.2857$ accidents in one day, and the poisson rv would be $1$, so $X \sim Po(0.2857)$, making the equation --
$$P(X = 1) = \frac{e^{-0.2857} 0.2857^1}{1!} = 0.2147$$
Is my thinking correct? should I be looking at these problems from a different perspective?