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Re: Conditional Probability

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**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 $\displaystyle (a,b)$ is a pair that tells us that person A arrives at time a, $\displaystyle 0\le a\le 30$ and person B arrives at time b, $\displaystyle 0\le b\le 30$. They meet if $\displaystyle |a-b|\le 5$.

Find the area in that strip inside the square.

Re: Conditional Probability

thank you very much for you help :D

the diagram was especially helpful ^^

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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

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

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

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