Time problem probability 
The bus stops randomly between 6:55 and 7:05 and the student comes on the station somewhere between 7:00 and 7:07. Whats the probability that student will catch the bus?

The answer is $5/28$, but I don't understand why, since we observe time between 6:55 and 7:07 and that is 12 min and the student will catch the bus if he comes somewhere between 7:00 and 7:05. So why isn't the correct answer $5/12$?
 A: For the student to catch the bus:


*

*The student must arrive between 7:00 and 7:05: probability ${\large{\frac{5}{7}}}$.

*The bus must arrive between 7:00 and 7:05: probability ${\large{\frac{1}{2}}}$.

*and, the student must arrive before the bus: probability ${\large{\frac{1}{2}}}$.


The product yields ${\large{\frac{5}{28}}}$.

Alternatively, we can model it in the $xy$-plane, as follows . . .

Let $x$ be the number of minutes after 6:55 that the student arrives, and let $y$ be the number of minutes after 6:55 that the bus arrives. Then the sample space is the rectangular region 
$$5 \le x \le 12$$
$$0 \le y \le 10$$
The student catches the bus if and only if $x \le y$.

So find the part of the rectangle such that $x \le y$, (graph it, then find the area by elementary geometry), and divide by the area of the rectangle.

The area of the rectangle is $(7)(10)=70$.

The part of the rectangle such that $x \le y$ is an isosceles right triangle with legs of length $5$, which has area ${\large{\frac{25}{2}}}$. 

Thus, the desired probability is 
$$\frac{(\frac{25}{2})}{70} = \frac{5}{28}$$
A: For any instant in time $t$ between 7:00 and 7:05,
$dP(t_{\text{student}} ≤ t_{\text{bus}}) = (\text{chance student arrives at $t$}) \cdot (\text{chance bus arrives after $t$})$
$dP(t_{\text{student}} ≤ t_{\text{bus}}) = \dfrac{dt}{7} \cdot \dfrac{(5-t)}{10}$
Integrating,
$P(t_{\text{student}} ≤ t_{\text{bus}}) = \dfrac{[5t - t^2/2]_{0}^{5}}{70}$
$P(t_{\text{student}} ≤ t_{\text{bus}}) = \dfrac{25}{140}$
$P(t_{\text{student}} ≤ t_{\text{bus}}) = \dfrac{5}{28}$
