Probability question?

Question

One couple meet each other between 2pm to 3pm with an understanding that each will wait no longer than 15 mins!. What is the prbability that they will meet???

Answer 1

Let
x = arrival time within the hour, where x varies from 0 to 1.
p = probabililty of meeting

p/2 = ∫{(1/4) + x}dx (from 0 to 1/4) + ∫(1/2)dx (from 1/4 to 1/2)
= (x/4 + x²/2) | (0 to 1/4) + x/2 | (1/4 to 1/2)
= (1/16 + 1/32 - 0 - 0) + (1/4 - 1/8) = 3/32 + 1/8 = 7/32

By symmetry the last half of the hour is a minimum of the first.

p = 2(p/2) = 2(7/32) = 7/16

The probability that they meet is 7/16.

Answer 2

7/16

The easiest way to get plot the arrival times in the x,y plane. The possible arrival times are the square defined by 2<=x<=3 and 2<=y<=3. For the couple to meet, we must have x>y-.25 and x
Edit: Northstar did the integral method. Cool! By the way, you can see it is somewhere between 1/4 and 1/2, but just noting that if one person arrives at 2pm, the probability of a meeting is 1/4, but if one person arrives at 2:30pm, the probability of a meeting is 1/2.

Answer 3

( I assume that each person has no predetermined
strategy and arrives at the meeting at a random time between
2pm and 3pm)

Let a = how many hours after 2 pm the first person arrives
Range of a = [0,1] inclusive, real numbers.

Let b = how many hours after 2p the second person arrives
Ranges of b = [0,1] inclusive, real numbers.

a - b has range [-1, +1]

For them to meet, a - b must be [-0.25 , +0.25]

I'd guess the probability is 25%.

Answer 4

It's one hour -> 60 mins

Since they no longer wait more than 15 mins, it's going to be 15/60 -> 1/4 -> 0.25