Buses arrive at a bus stop at 15 min intervals starting at 7 am assume that a passenger arrives at the bus stop at random time X (given in minutes after 7 am ) with PDF
$$f_X(x) =
\begin{cases}
\frac {x(60-x)}{36000}, & \text{if $ 0\le x \le 60 $} \\
0, & \text{otherwise}
\end{cases}$$
I have try to set everything as the following :
$X$: time the passenger arrive at the bus stop
$Y$: the waiting time
Since I need the passenger to be waiting more than 10min so I have $P(Y>10)$
With $P(15-X>10) = P(X<5)=P(X \le 4)= \int_{0}^{4} \frac {x(60-x)}{36000}dx=0.012=1.2$%$ $
I was wondering if that can be a viable solution and if for any reasons there's a new way to proceed please let me know.
Best Answer
Guide:
If someone has to wait for more than $10$ minutes, his corresponding arrival time in term of $x$ is
$$(0,5) \cup (15, 20) \cup (30, 35) \cup (45, 50)$$