Time Between Events Probability Question

[Patrick D. Rockwell]

Ok, let's say that 3 people agree to meet between 3 and 4 P.M. Each of
them is willing
to wait 15 minutes. The probability that NONE of them will meet were
d = .25 =
(15 minutes)/(1 hour) and n = 3 is

(1-d(n-1))^3 = (1-(0.25)*2)^3 = (1-.5)^3 = .5^3=.125.

Now, what if during 3 and 4 P.M., a computer is going to emit a tone 3
times. The
tones are given at random times during 3 and 4 P.M.. What is the
probability that all 3
tones are at least 15 minutes apart? Is it .125 as above? Should I
multiply or divide by
3 factorial for this second problem?

Thanks in advance.

[cliff]

On May 14, 11:21 pm, "Patrick D. Rockwell" <prockw...@thegrid.net>
wrote:

For the first problem, let x,y,z be the the arrival times of persons
1, 2 and 3. Assume that they are independent and uniformly
distributed in [0,1]

The integral
(from 0 to 1) (from x+1/4 to 1) (from y+1/4 to 1) (1) dz dy dx = 1/96
Is the volume of all arrival times (x,y,z) with x+1/4 < y and y+1/4 <
z.
This takes care of the probability that they all miss each other, if x
< y < z.
Since there are 5 other possibilities of ordering x,y,z, the total
probability is 6/96 = 1/16.

So I get a different result than you.

The second problem is equivalent, if you imagine the beeps being
produced by three separate alarms, 1,2 and 3, that are independently
and uniformly distributed in [0,1]. The 1st beep corresponds to the
arrival of the person 1, etc.