Little math problem(s) October 2020

[Albert Chan]

(10-09-2020 07:56 PM)Albert Chan Wrote:  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

This assumed all trains stayed at the station for 5 minutes.
In other words, the station were never empty.

More realistic situtation is train only stay relative short time, then leave. Say 1 minute.

P(25 ≤ x ≤ 31) = (y at x=28) * (6 minutes) = 0.03833

This slight adjustment meant chance of both on the same train dropped to 1/26

Recalculate, we have daily probability of encounter = 1/(26 * 8 * 25) = 1/5200
Chance of never seating next to each other (for the year) = (1-1/5200)^240 ≈ 95.5%

At least once for the year seating next to each other = 1 - 95.5% = 4.5%