The bus arrives at the station every $45$ minutes in the time interval from $7am$ to $7pm$. The person at the station comes in order. The time that the person will wait is uniformly distributed in this time interval.
$a)$ graphically represent this distribution
$b)$ what is the probability that the random person will wait more than $10$ minutes
$c)$ what is the probability that the random person will wait in the interval of (10-20min)
My attemp is: let $X$ its time that the random person comes in station, $X\thicksim U(0,720)$
$a)$
but I don't know if the graph is good, or if something else needs to be added
$b)$ $$P(X\in(0,5)\cup(15,20))=P(0<X<5)+P(15<X<20)=\int_0^5\frac{dx}{720}+\int_{15}^{20}\frac{dx}{720}=\frac{1}{72}$$ but I don't know if the solution is good
$c)$ $$P(X\in (10-20min))=P(10<X<20)=\int_{10}^{20}\frac{dx}{720}=\frac{1}{72}$$ but I don't know if the solution is good.
Help me please. Thanky very much for my hard
The bus comes at the times of $45i$ for $i=0,1,\cdots ,15$ in the time interval $[0,720]$ (note that $720=45\times 16$), therefore the probability of b) becomes $$P=\Pr \left\{\bigcup_{i=0}^n45i<X<45(i+1)-10\right\}=\sum_{i=1}^n\Pr\{45i<X<45i+35\}$$
For part c) similarly that is $$P=\Pr \left\{\bigcup_{i=0}^n45(i+1)-20<X<45(i+1)-10\right\}=\sum_{i=1}^n\Pr\{45i+25<X<45i+35\}$$