A country bus driver picks up passengers randomly and independently at a mean rate of $\ 12$ per hour.
The time at which he picks up his first passenger is $\ T$ hours. Explain why
$$\ P(T < t) = 1 − e^{−12t} ~for~ t > 0.$$
Attempt
Let's say the driver picks his first passenger at $\ t$ hours. Then it is reasonable to say that before $\ t$ hours he had $\ 0$ passengers. $$\ P(T<t) = 1-Po(12t) $$ $$\ P(T<t) = 1-e^{-12t}\frac{(12t)^0}{0!} $$ $$\ P(T<t) = 1-e^{-12t} $$ $\ t$ has to be greater than $\ 0$ since negative time doesn't exist (at least in this context).
So$$\ P(T < t) = 1 − e^{−12t} ~for~ t > 0.$$
Is this explanation correct? If not, can somebody explain why?
