1. ## Conditional Probability

This is one of the practice questions in my extension class,
Apologies if the title topic is wrong, I'm a bit confused on how to approach the question:

Two friends agree to meet in the town square between 1:00pm and 1:30pm. They agree that they will wait for 5 minutes after arriving. If the other person does not arrive during that five minutes, the first person will leave. What is the probability that they meet?

So far I'm thinking that the answer is something like $\frac{(number of actual outcomes)}{30^2}$
but I don't know how to find the number of actual outcomes or if this is even how to solve the question.

any help is much appreciated, thanks in advance.

2. ## Re: Conditional Probability

Originally Posted by angelaiyes
Two friends agree to meet in the town square between 1:00pm and 1:30pm. They agree that they will wait for 5 minutes after arriving. If the other person does not arrive during that five minutes, the first person will leave. What is the probability that they meet
Think of a 30x30 square. If $(a,b)$ is a pair that tells us that person A arrives at time a, $0\le a\le 30$ and person B arrives at time b, $0\le b\le 30$. They meet if $|a-b|\le 5$.
Find the area in that strip inside the square.

3. ## Re: Conditional Probability

thank you very much for you help
the diagram was especially helpful ^^

4. ## Re: Conditional Probability

I don't know if you will like it,
but here is my alternative way of thinking this through.

Divide the half-hour into 6 five-minute intervals.
In the attachment, if the first person arrives at any time from C to E,
then the friend has a 10-minute interval in which to arrive such that the 2 friends meet.

Hence, in that region from C to E, the probability they meet is 1/3.
In the first and last 5-minute intervals of the half-hour,
the probability they meet is less than 1/3.
This is because the 2nd friend's "window" varies linearly from 5 to 10 minutes
depending on exactly where in the 1st and last intervals from A to B and from E to F
that the first person arrives.

Let's say the first person arrives at A.
Then the 2nd friend must arrive somewhere from A to B.
The probability of the 2nd friend arriving in this interval is 1/6.

Hence, if the first person arrives at B,
anywhere from A to C,
then the probability of the friends meeting varies linearly from 1/6 to 1/3.

If the 2nd person arrives at F, then again, there is a 1/6 probability the friends meet,
as the friend may have arrived at any time between E and F and will have been waiting.

We can calculate the probability the friends meet by calculating the fraction of the 1x1 square
underneath the purple line in the attachment.
This is

$\frac{4}{6}\frac{1}{3}+\frac{3}{4}\; \frac{1}{6}\frac{1}{3}(2)$

$=\frac{4}{18}+\frac{3}{2}\;\frac{1}{18}$

$=\frac{8}{36}+\frac{3}{36}$