Probability that a passenger will catch the bus A bus and a passenger arrive at a bus stop at a uniformly distributed time over the interval $0$ to $1$ hour. Assume the arrival times of the bus and passenger are independent of one another and that the passenger will wait up to $15$ minutes for the bus to arrive. What is the probability that the passenger will catch the bus?
Hint: Let $Y1$ denote the bus arrival time and $Y2$ the passenger arrival time. Determine the joint density of $Y1$ and $Y2$ and determine $$P(Y2 \leq Y1 \leq Y2+\frac{1}{4}).$$
We already know that the answer to this question is $\frac{7}{32}$. Any help would be much appreciated.

There is a $\frac{3}{4}$ chance the bus will be later than 10:15. Therefore there is a $\frac{1}{4}$ of an hour window of the person $Y2$, arriving on time. If $Y1 < \frac{1}{4}$ and $Y2 > \frac{1}{4}$ the probability of the person missing the bus is certainly $1$. 
If $Y1 (\frac{1}{4})$ AND $Y2 (\frac{1}{4})$ arrives before 10:15, then $\frac{1}{4}+\frac{1}{4} = \frac{1}{2}$ chance of catching the bus. 
So we have: $$\frac{3}{4} \cdot \frac{1}{4} + (\frac{1}{4}\cdot \frac{1}{4}  \cdot \frac{1}{2}) = \dfrac{3}{16} +  \dfrac{1}{32} =  \dfrac{7}{32} $$
