# Exercise 4.1Z: Appointment to Breakfast

Ms. M. and Mr. S. are known to meet often for a joint breakfast:

• Both promise to come to such a meeting on a certain day between 8 am and 9 am.
• Further,  they agree that each of them will arrive in this period  (and only in this period)  on  "good luck"
and wait up to fifteen minutes for the other.

Hints:

• The exercise belongs to the chapter  Two-Dimensional Random Variables.
• Use the minute of arrival time as the time in the following questions:
"Minute = 0"  stands for 8 o'clock, "Minute = 60"  for 9 o'clock.
• The exercise arose before the 2002 German Bundestag elections,  when both Dr. Angela Merkel and Dr. Edmund Stoiber wanted to become the CDU/CSU's candidate for chancellor.
• At a joint breakfast in Wolfratshausen,  Ms. Merkel renounced.  The later election was won by Gerhard Schröder  (SPD).

### Questions

1

What is the probability  $p_1$  that the two will meet when Mr. S. arrives at 8:30?  Give reasons for your answer.

 $p_1 \ = \$ $\ \%$

2

Which arrival time should Ms. M. choose if she does not actually want to meet Mr. S.,  but still wants to keep to the agreement made?
What is the probability  $p_2$  that Ms. M. and Mr. S. will meet?

 $p_2 \ = \$ $\ \%$

3

Which arrival time should Ms. M. choose if she not only wants to avoid a meeting as much as possible,  but also wants to minimize the waiting time?

 $\rm minute \ = \$

4

What is the probability  $p_4$  for a meeting in general,  that is,  if both actually appear on "good luck"?

 $p_4 \ = \$ $\ \%$

### Solution

#### Solution

(1)  If Mr. S. arrives at 8:30,  he will meet Ms. M. if she arrives between 8:15 and 8:45.  Thus the probability:

$$p_1 = \text{Pr(Mr. S. meets Ms. M.)}\hspace{0.15cm}\underline{=50\%}.$$

(2)  If Ms. M. arrives at 8 a.m.,  she meets Mr. S. only if he arrives before 8:15.

• If Ms. M. arrives at 9 a.m.,  Mr. S. must arrive after 8:45 a.m. so that they can meet.
• The probability of meeting is the same in both cases:
$$p_2 = \big[\text{Min Pr(Mr. S. meets Ms. M.)}\big]\hspace{0.15cm}\underline{=25\%}.$$

(3)  Of the two arrival times calculated in  (2),  9 o'clock  $(\underline{\text{Minute = 60}})$  is more favorable,
since she – if Mr. S. is not there – can leave immediately.

(4)  The probability  $p_4$  is given as the ratio of the red area in the graph to the total area  $1$.

• Using the triangular areas,  one obtains:
$$p_4=\rm 1-2\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}=\frac{7}{16}\hspace{0.15cm}\underline{=\rm 43.75\%}.$$