Lunch Meeting Probability for two person to meet in given 1 hour slot and none would wait more then 15 minute. Two friends who have unpredictable lunch hours agree to meet for lunch at their favorite restaurant whenever possible. Neither wishes to eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1 pm, and each will wait for the other for 15 minutes or until 1:00. What is the probability that the friends will meet for lunch on a given day?
Per @AppDeveloper's request changing it from a comment to an answer:

Just and idea: consider $0$ to $30$ min, the other half is the same by symmetry. 

If A arrives at $0$ min, B has to arrive between $0$ and $15$ min, i.e., $p(B\leq 15|A=0)=\frac{1}{4}$. 

If A arrives after $15$ min, $p(B|A)=\frac{1}{2}$. 

Applying conditional probabilities and integrating, get for $t \geq 15$ min $p(B|A)p(A)=\frac{1}{8}$ and for $0 \leq t \leq 15$,  $p(B|A)p(A)=\frac{3}{32}$ Adding together and multiplying by $2$, get $\frac{7}{16}$.
No matters who arrives first.
Assume $A$ arrives first. 2 cases then to be considered:
Case 1: $A$ arrives between 12.00-12.45. 
Then the chance that both will meet equals: $3/4 \cdot 1/4 = 3/16$.
Case 2: A arrives between 12.45-13.00.
Then the chance that both will meet equals: $1/4\cdot 1= 1/4$.
Chance that both will meet then equals: $3/16 + 1/4 = 7/16$.