00:01
The asker question support that a man leaves for work between 8 a .m.
00:09
And 8 .30 a .m.
00:17
And takes between 40 and 50 minutes to get to the office.
00:25
Between 40 and 50 minutes to get to the office.
00:34
Let x denote the time of departure and let y denote the time of travel.
00:42
If we assume that these random variables are independent and uniformly distribution, it's required to find the priority that he arrives at the office before 9 a .m.
01:00
Here i prefer to switch all the runner variables to have a units of minutes.
01:07
Then if we start our time at 8 a .m., this means that x follows a uniform distribution between 0 and 30 minutes, while y follows an uniform distribution between 40 and 50 minutes and we can say that now z is the time of 9 a .m.
01:44
Minus 8 a .m.
01:46
Or just z.
01:49
And we want to find the probability of z to be less than 60 minutes.
01:59
This is the same as before 9 a .m.
02:02
Because we have started our 0 at 8 a .m.
02:11
How we can approach this problem we have many ways.
02:15
The easiest way is to construct the random variable x here and the random variable y here...