Date problem with asymmetric waiting time

Question

Two people are going on a blind coffee date, one of them will show up randomly at a time between 12:00 and 12:45, and stay for 30 minutes before leaving. The other will show up randomly at a time between 12:30 and 13:00 , and stay for 15 minutes before leaving. What is the probability that the two actually meet?

Numerically I got an answer of about $67$%. I am aware of the solution for the case of symmetrical waiting time and arrival times, but I am unsure how to generalize it in this case.

Answer 1

To make the solution cleaner, I will scale the time such that "$0 \cong 12:00$" and "$1\cong 13:00$". I will also assume that the arrival times are independent:

Let $X$ be the arrival time of the first person and $Y$ be the arrival time of the second person. Since we're dealing with uniform distributions, $X \sim U\left([0,\frac{3}{4}]\right)$ and $Y \sim U\left([\frac{1}{2},1]\right)$. Geometrically, graphing the density function of $(X,Y)$, we therefore get a "rectangle" of constant height.

Now to the interesting bit: The first person is in the coffee shop during the time interval $[X, X + \frac{1}{2}]$ and the second person during the time interval $[Y, Y + \frac{1}{4}]$. Asking for them to meet, is equivalent to $[X, X + \frac{1}{2}] \cap [Y, Y + \frac{1}{4}] \neq \emptyset$ (i.e. the time intervals of the two intersecting), so

1. $X + \frac{1}{2} \geq Y$ and
2. $Y + \frac{1}{4} \geq X \iff Y \geq X - \frac{1}{4}$.

Again, as we're dealing with uniform distributions, the probability that they'll meet is given by the "portion of the rectangle" for which 1. (in red) and 2. (in purple) holds.

That the portion is exactly $\frac{2}{3}$, should be clear. :)

Answer 2

The only way that the two can miss each other is if the 1st person leaves before the 2nd person gets there.

If the 2nd person arrives at 12:30 + $t$ (minutes) then the chances of a miss are $\frac{t}{45}$.

Therefore, the chance of a miss are

$$\frac{\int_0^{30} \frac{t}{45} dt}{\int_0^{30} (1)dt} = \frac{\frac{30^2}{2 \times 45}}{30} = \frac{1}{3} .\tag1$$

In (1) above, the denominator is used to normalize the probability represented by the numerator.

Addendum
As suggested by the answer of MXXZ, an alternative approach would be to reason that with respect to the probability function referred to in (1) above, the function is linear.

Therefore, you can reason that the overall probability of a miss will equal the average probability of $\displaystyle \left(\frac{t}{45}\right)$, as $t$ ranges from $(0)$ through $(30)$.

Since the probability function is linear, the average value will occur when $t$ is at the midpoint of the interval $[0,30].$ Therefore, the overall chance of a miss must be $\displaystyle \left(\frac{15}{45}\right)$.