amacorrit80

## Answered question

2022-07-19

Geometric Probability in Multiple Dimensions
I understand how Geometric Probability works in 1, 2, and 3 dimensions, but is it possible to do these problems in, say, 5 dimensions? For example,
Five friends are to show up at a party from 1:00 to 2:00 and are to stay for 6 minutes each. What is the probability that there exists a point in time such that all of the 5 friends meet?

### Answer & Explanation

Izabelle Frost

Beginner2022-07-20Added 13 answers

Step 1
Consider the simpler problem of 2 people. Chart on the x axis the time of stay of first person. Chart on the y axis the time of stay of second person. The cartesian product of these intervals together form a rectangle in ${\mathbb{R}}^{2}$
Question: When do the two meet, and what does this mean geometrically? It's easy to see that whenever this rectangle intersects the $x=y$ line, then the two people meet (or have an overlapping time), or else if the rectangle is fully below $x=y$, or fully above they don't meet.
Extending to higher dimensions: When formulated this way, it should be the case that in ${\mathbb{R}}^{n}$, when the rectangle formed by the cartesian product of the intervals of stay of different people intersects the line ${x}_{1}={x}_{2}=\cdots ={x}_{n}$, (or in parametric form, $\lambda \left[1,1,\dots ,1\right]\phantom{\rule{1em}{0ex}}\mathrm{\forall }\lambda$) then, all the people have some overlapping time of stay.
Step 2
Probability of overlap: The probability is now the measure of the union of all the rectangles (or hypercubes in ${\mathbb{R}}^{n}$) intersecting with the line (${x}_{1}={x}_{2}=\dots$) vs the measure of the entire space under consideration, say under normalization, $\left[0,1\right]×\left[0,1\right]×\dots \left[0,1\right]$

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