Laurel appears at Comedy Cafe at a random time between 1:00 and 2:00. Hardy turns up at a random time between 2:00 and 2:30. The probability that Laurel waits for no more than 40 minutes for Hardy can be expressed in the form m/n, where m and n are relatively prime positive integers. Find m + n.

Answer 1

Events:
S=Laurel waits 40 minutes or less for hardy.
~S=Laurel waits more than 40 minutes.

Assumptions: both arrivals are random and create a uniform probability distribution.

Between 1:00 and 1:20, Laurel will be sure to wait 40 minutes or more, since Hardy does not arrive before 2:00. So P(S)=0

Between 1:50 and 2:00, Laurel will always wait 40 minutes or less for Hardy, so P(S)=1.

Between 1:20 and 1:50, the probability of success increases from 0 to 1, i.e. this is the period where the 40 minutes covers more and more of the half-hour that Hardy arrives.

If we sum the three areas and divide by the duration (60 minutes), we have the probability that Laurel would have waited 40 minutes or less for Hardy.

A=0*20+(1/2)(0+1)*30+10*1
=0+15+10
=25

Over the one hour Laurel could arrive,
P(S)=25/60=5/12
=>
m=5, n=12, m+n=5+12=17

1 .....
* .
* .
* .
0........

A sketch of the probability distribution.

Answer 2

Sorry the sketch did not work out! :(