Little math problem(s) October 2020

[Albert Chan]

Differences of 2 uniform random distribution variables followed triangular distribution.

Let x = time differences (minutes) between 2 person arriving to the same station,
triangular distribution curve peaked when x = 0, drops to 0 if |x| ≥ 120
(2 hours window per day where encountering is possible)

If |x| ≤ 120, y = 1/120 - |x|/120²
If |x| > 120, y = 0

Assuming 5 or 6 stations about a difference of 30 minutes:

P(|x-30| ≤ 5) = P(25 ≤ x ≤ 35) = (y at x=30) * (10 minutes) = 0.0625

Chance both will be on the same train ≈ 1/16
Chance both will be on the same car ≈ 1/8
Chance both seat next to each other ≈ 2/47 ≈ 1/25 (account for corner seats)

Chance of all of above to happen = 1/(16 * 8 * 25) = 1/3200

Assuming 240 days a year both will be using the train.
Chance of never seating next to each other (for the year) = (1-1/3200)^240 ≈ 93%

Its complement, at least once for the year seating next to each other = 1 - 93% = 7%

This might be completely wrong ...