Question

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:30 and 12:10 and if the woman independently arrives at a time uniformly distributed between 11:50 and 12:35, what is the probability that the first to arrive waits no longer than 15 minutes?

Answer 1

5/21

Explanation:

From the question and time interval given the Man will most likely arrive before the woman so i will breakdown the various possible arrival time of the Man in relation to the woman's arrival time.

lets say

11:40 to 11:50

probability of arrival by the man = 10/35

during this period the woman has a probability of arriving less than 10 minutes within a 5 minutes interval

her own probability = 5/60

11:50 to 12:00

probability of arrival by the man = 10/35

during this period the woman has a probability of arriving within 10 minutes of the man's arrival within a 15 minute interval

her probability of arrival = 15/60

12:00 to 12:15

probability of arrival by the man = 15/35

during this period the woman has a probability of arriving within 20 minutes interval

her probability = 20/60

therefore to get the probability that the first to arrives waits no longer than 15 minutes

we multiply the various probabilities for each time interval and then add them up

= ( 10/35 * 5/60 ) + ( 10/35 * 15/60 ) + ( 15/35 * 20/60 )

= 500/2100

= 5/21

Answer 2

The probability that the first to arrive waits no longer than 15 minutes is 1/3

Explanation:

Let X be the distribution of time where the man arrives and Y the distribution for the woman. Since it is a 40 minute period, the density function of X is

`f_X(t) = 1/40 \, I_([11:30, 12:10])`

(Note: it is weird not to work with real numbers, however, you can think an hour as 60 minutes; for example 11:30 = 11*60+30 = 690, so we are working with real numbers after all. But it is more comfortable to work with minutes and hour notation)

The density of Y is

`f_Y(t) = 1/45 \, I_([11:50, 12:35])`

Note that if the man arrives at 11:30, he will have to wait more than 15 minutes regardless of when the woman arrives; he has to arrive at least 11:35 for a chance. The same way, no matter how late the man arrives, he will wait more than 15 minutes if the woman arrives after 12:25

We can obtain the desired probability spltting the event in 2: when the man arrives earlier and when the woman does. For the first even we need to integrate the product of the density functions (which are independent), with X taking any value greater than 11:35 until 12:10 and Y being between X and X+15. Then we integrate with X between 11:50 + 15 = 12:05 and 12:10 and Y between X-15 and X

`P(|X-Y| \leq 15) = P(0 \leq X-Y \leq 15) + P(0 \leq Y-X \leq 15)`

Now,

`P(0 \leq X-Y \leq 15) = \int\limits^(12:10)_(12:05)\int\limits^(x)_(x-15) {(1)/(45*40)} \, dy \, dx = \int\limits^(12:10)_(12:05) (x - (x-15))/(1800) \, dx = \\((12*60+10) - (12*60+5))/(120) = 5/120 = 1/24`

And,

`P(0 \leq Y-X \leq 15) = \int\limits^(12:10)_(11:35)\int\limits^(x+15)_(x) {(1)/(1800)} \, dy \, dx = \int\limits^(12:10)_(11:35) ((x+15) - x)/(1800) \, dx = \\((12*60+10) - (11*60+35))/(120) = 35/120 = 7/24`

Thus, P(|X-Y| ≤ 15) = 1/24 + 7/24 = 1/3.

The probability that the first to arrive waits no longer than 15 minutes is 1/3.