Here is my Question:
A country bus driver picks up passengers randomly and independently at a mean rate of 12 per hour.
(i)Find, correct to 3 decimal places, the probability that he picks up at least 3 passengers in a period of 15 minutes
(ii) Show that the probability that he picks up at least one passenger in a period of t hours is
$1-e^{-12t}$
(iii) The time at which he picks up his first passenger is T hours. Explain why $P(T<t)=1−e^{−12t}$ for $t>0$.
My Attempt
(i) 12 is per hour, so therefore $\lambda=3$ for 15 mins.
Atleast 3 means
$P(X\geq3)=1-(P(X=1)+P(X=2))$
Therfore using the formula $P(X=x)=\frac{e^{-\lambda}(\lambda)^x}{x!}$
$=1-(\frac{e^{-3}(3)}{1!}+\frac{e^{-3}(3)^2}{2!})$
But this is wrong, the correct answer comes when doing $1-(P(X=1)+P(X=2)+P(X=3)$, why is that?
(ii)(iii) I dont know how to get this for t, I dont know how to start with to show an attempt, please help.
(i) First, as the author did, we know that 12 is the average per hour so 3 is the average in a quarter of an hour. Now, the complement of at least three is $0,1,2$ so, $$ P(\textrm{at least 3})=1-[P(X=0)+P(X=1)+P(X=2)]=1-e^{-3}-3e^{-1} -\frac{e^{-2} 3^2}{2!} $$ (ii) We know that 12 is the average per hour so $12t$ is the average in $t$ hours. Now, the complement of at least one is $0$ so, $$ P(\textrm{at least 1})=1-P(X=0)=1-e^{-12t} $$ (iii) Let $T$ be the time the first passenger is picked up. The first passenger is picked up before $t$ if and only if at least one passenger were picked up by time $t$, which according to part (ii) is $P(T<t)=P(\textrm{at least one by $t$})= 1-e^{-12t}$.