TITLE: Combined probability distribution function of two uniformly distributed random variable.
QUESTION [1 upvotes]: I was solving a uniform probability distribution question, the question was : $\\$ probability of meeting two people A and B between 1 PM and 2 PM, if none of them waits more than 15 minutes for the other. Given that probability of arrival is uniform between 1 PM to 2 PM for both of them.
I took the time span as L and separated two zones. [0,3L/4] and [3L/4,L].
In the first zone if A comes first at time x B has the option to come anytime in [x,x+L/4].
In the second zone,if A comes first at x then B must come in [x,L-x].
I did the integration in two zones and found P(A-B) = 7/32,and multiplied by 2 gives the total probability.
Later I framed (visualized) the same question as follows :
In Cartesian co-ordinate system,along the x axis two points p and q are selected uniformly at random in $\left [ 0,L \right ]$ where L > 0.What is the probability of $\text{dist(p,q)} \leq \frac{L}{4}$.
Then I modified it to two dimensional geometry problem :
In the Cartesian plane, selection of a point P along the y axis in [0,2] is uniformly random. Similarly selection of a point Q along the x axis in [0,2] also uniformly distributed. What is the probability of the area of the triangle POQ to be less than or equal to 1, where O is the origin ?
I solved it using integration in two parts:

if x is in [0,1] then y is allowed to have the full range of [0,2]
if x is in [1,2] then y is allowed to have the range [0,2/x]

I got probability as 0.846.
My questions are:

Is it correct ?
What is the more generic way to to such problem ? In both the above cases I drew some geometry and did some integration. In the second problem, we are interested in the quantity area (xy/2). But what if we are given a random expression like (3xy+x^2). Please elaborate any other method.

REPLY [1 votes]: The event $x$ and $y$ are independent and uniformly distributed
so that geometrically the event space is represented by a square $1\;h\; \times \,1\;h$
of uniform density.
The condition for the meeting to take place is.
$$
\left| {y - x} \right| \leqslant 1/4
$$
i.e.
$$
0 \leqslant y - x \leqslant 1/4\quad  \vee \quad  - 1/4 \leqslant y - x < 0
$$
or
$$
x \leqslant y \leqslant x + 1/4\quad  \vee \quad x - 1/4 \leqslant y < x
$$
which is the area in yellow in the sketch.
 
The probability of meeting is easily calculable as:
$$
P_{\text{meet}}  = 1 - 2\frac{1}
{2}\left( {\frac{3}
{4}} \right)^2  = \frac{7}
{{16}}
$$
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