Question

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10 and if the woman independently arrives at a time uniformly distributed between 11:45 and 12:20, what is the probability that the first to arrive waits no longer than 5 minutes?

Answer 1

Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$