Don't be late!!

Probability Level 2

Two friends plan to meet in a library over a period of one hour. Their arrival times are independent and distributed randomly over that hour period. Both of them agree to wait 15 minutes (or until the end of the agreed hour, in case the arrival time is less than 15 minutes from it). If the other doesn't arrive during that time, they leave.

As examples, if they agree to meet between 5 and 6:

If one of them arrives at 5:17, she waits until 5:32.
If one arrives at 5:51, she waits until 6:00.
If one arrives at 5:23 and the other arrives at 5:35, they meet.
If one arrives at 5:13 and the other arrives at 5:49, they don't meet.

What's the probability of both friends meeting? Write your answer up to 4 decimal places.

The answer is 0.4375.

1 solution

Miguel Vásquez Vega
Sep 19, 2018

Let $t_1$ and $t_2$ be the arrival times of each friend. We know that they'll meet if $|t_1-t_2|=15$ minutes, or, the same, if

$t_1-t_2>15\text{ minutes}\qquad\Longrightarrow\qquad t_2>t_1+15\text{ minutes}$ and $-15\text{ minutes}<t_1-t_2\qquad\Longrightarrow\qquad t_2<t_1-15\text{ minutes}.$

If we plot this regions on a $t_1-t_2$ plane as shown, we obtain the time regions in which the friends can meet.

Hence, from all the possible pairs of time arrivals they've got, given by the 60 minutes side square, $60^2$ , they can meet in $60^2-45^2$ of them. This is, the probability $P(E)$ of both friends meeting is

$\displaystyle \boxed{P(E)=\frac{60^2-45^2}{60^2}=\frac{7}{16}=0\text{.}4375=43\text{.}75\%.}$

Don't be late!!

The answer is 0.4375.

1 solution

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