The time until the next car accident for a particular driver is
exponentially distributed with a mean of 200 days. Calculate the
probability that the driver has no accidents in the next 365 days, but
then has at least one accident in the 365-day period that follows this
initial 365-day period.
Attempt
Let $T$ be the time it takes for a driver to have a car accident. We are given $T$ is $exp( \lambda = 1/200 )$. We need to find
$$ P(T > 365) = 1 – F(365) = 1 – 1 + e^{-365/200} = 0.1612 $$
Is this correct? MY answer key says the correct answer should be $\boxed{0.1352}$. What am I missing here?
Best Answer
You want the first accident to be between the first year and second year.
\begin{align} P(365< T \leq 2 \cdot 365) &= F(2 \cdot 365) - F(365) \end{align}