Coffee Shop Meeting $A$ and $B$ decide to meet at a cafe between $5$ p.m. and $6$ p.m. They agree that the person who arrives first at the cafe would wait for exactly $15$ minutes for the other. If each of them arrives at a random time between $5$ p.m. and $6$ p.m., what is the probability that the meeting takes place?
I figured that if one of them arrive at the first minute then the probability of the two meeting each other would be $15/60$, because the second person could arrive from the $1^{st}$ minute till the $15^{th}$ minute and meet with him. Similarly if the first person arrives at the second minute the probability would be $16/60$. This will go on till the $14^{th}$ minute and the probability would be $29/60$. The probability will remain $29/60$ till the $45^{th}$ minute, after which it will gradually decrease in the order $28/60, 27/60,... , 15/60.$
I am not sure if my approach is correct. Also I am stuck after a point with my approach. Please explain elaborately how to solve such questions.

A graphical solution to this problem:
Let A and B be Alice and Bob's arrival times, both variables taking on values between 0 and 1.
Since Alice and Bob must arrive within 1/4 hours of each other, the following equation must be satisfied:
$$|A - B| \leq 1/4$$
We can break apart this absolute value into the following two cases:
$$ A \geq B + 1/4, and, A \leq B - 1/4$$
Graphing these inequalities on a plot where A and B are between 0 and 1, creates three regions. The region where the inequality is satisfied is the diagonal strip down the middle.
Since probability is proportional to area, and the total area of the plot is 1, the area of the diagonal strip is the probability that the meeting will take place. Calculating that probability:
$$ P = 1 - (3/4)^2 = 7/16$$
