# Thread: Meeting Time Probability

1. ## Meeting Time Probability

Hey guys, im not the greatest at probability but here is a puzzle which i cant get....any idea!:

"We have 2 indivduals who agree to meet at a location between NOON and 1 P.M. They also agree to wait for 15 minutes for the other to show up (after that they will leave).

If each of them selects the time of arrival at random, what probability is it that the meeting will actually happen?"

2. If each of them selects the time of arrival at random, what probability is it that the meeting will actually happen?"
I assume you mean uniform distribution over $\displaystyle [0,1]$.

Let $\displaystyle X,Y$ be the time in hours after Noon that the first and the second person (resp) arrive.

You want to calculate $\displaystyle \mathbf{P}(|X-Y|<1/4)$

$\displaystyle \mathbf{P}(|X-Y|<1/4) = \mathbf{E}[\mathbf{P}(|X-Y|<1/4)\ |\ Y]$

$\displaystyle \int_0^1 \mathbf{P}(|X-y|<1/4)\ 1\cdot dy$

$\displaystyle \int_0^1 \mathbf{P}(y-1/4<X<1/4+y)dy$

Break into 3 regions, because if $\displaystyle Y<1/4\text{ or }Y >3/4$ then the lower/upper (resp.) bound on when $\displaystyle X$ is allowed to occur is bounded by 0/1 (resp.) rather than $\displaystyle Y-1/4\text{ or }Y+1/4$ (resp.).

$\displaystyle =\int_0^{1/4} \mathbf{P}(y-1/4<X<1/4+y) dy + \int_{1/4}^{3/4} \mathbf{P}(y-1/4<X<1/4+y) dy + \ldots$
$\displaystyle \ldots\int_{3/4}^1 \mathbf{P}(y-1/4<X<1/4+y) dy$

$\displaystyle =\int_0^{1/4} \int_{0}^{1/4+y} 1 dx dy + \int_{1/4}^{3/4} \int_{y-1/4}^{1/4+y} 1 dx dy + \int_{3/4}^1 \int_{y-1/4}^1 1 dx dy$

$\displaystyle =\int_0^{1/4} 1/4+y\ dy + \int_{1/4}^{3/4} 1/2\ dy + \int_{3/4}^1 5/4-y\ dy$

$\displaystyle =3/32 + 1/4 + 3/32=7/16$

Seems to me like they out to work out a different scheme.

3. Originally Posted by furnis1
Hey guys, im not the greatest at probability but here is a puzzle which i cant get....any idea!:

"We have 2 indivduals who agree to meet at a location between NOON and 1 P.M. They also agree to wait for 15 minutes for the other to show up (after that they will leave).

If each of them selects the time of arrival at random, what probability is it that the meeting will actually happen?"
Another approach (which boils down to the same thing as meymathis did):

Let X be the random variable number of minutes after noon that person X arrives. X ~ U(0, 60).

Let Y be the random variable number of minutes after noon that person Y arrives. Y ~ U(0, 60).

1. When $\displaystyle 0 \leq X \leq 15$ you require $\displaystyle 0 \leq Y \leq X + 15$ for the two people to meet.

2. When $\displaystyle 15 \leq X \leq 45$ you require $\displaystyle X - 15 \leq Y \leq X + 15$ for the two people to meet.

3. When $\displaystyle 45 \leq X \leq 60$ you require $\displaystyle X - 15 \leq Y \leq 60$ for the two people to meet.

On the XY-plane draw the square bounded by the lines X = 0, Y = 0, X = 60 and Y = 60. The area of this square is 3600 square minutes.

Cases 1-3 above define areas inside this square (unit is square minutes) for which the two people will meet:

1. 675/2.

2. 900.

3. 675/2.

Total area = 1350 square minutes.

Therefore the probability of the two people meeting is 1350/3600 = 7/16.

Note: This geometric approach only works because X and Y follow uniform distributions.