Hi, I'm a newbie to this. This is a genuine situation, whicjh I'm trying to mathematically solve. It should be easy for you guys, so I appreciate your help!
8 cars use a private drive to join a local road
Peak time for traffic leaving drive is 7-9am and 5-7pm
Assume 4 of cars are used for working nights though, so worst case is 4 cars out and 4 cars in between 7-9am and same between 5-7pm
Takes 15 secs to leave drive (i.e. be at bottom and look left, right and pull out)
Return takes 10 secs to check and pull in
In all likelihood, all 8 cars will work days and leave at the same time. However, worst case scenario (4 work nights, 4 work days), So……Work out probability for ONE day of cars meeting each other - one at bottom of drive and one wanting to turn in at same time
Leave: 4 cars * 15 secs = 1 min in a 120min period (7-9am)
Return: 4 cars * 10 secs = 40 secs in a 120min period (7-9am)
Leave: 4 cars * 15 secs = 1 min in a 120min period (5-7pm)
Return: 4 cars * 10 secs = 40 secs in a 120min period (5-7pm)
What is this for a month?
If you suss this, can you tell me the formula, so I can apply to other periods of the day.
My guess is the probability is (1/120)*(1/180) from 7am-9am. But this number seems far too small