Weyland R.

asked • 05/31/20

Very difficult probability problem More help would be appreciated! - What is the probability that at least one attendee is on the call with everyone else?

One Friday morning, suppose everyone in the U.S. (about 330 million people) joins a single Zoom meeting between 8 a.m. and 9 a.m. This being a virtual meeting, many people will join late and leave early.


In fact, the attendees all follow the same steps in determining when to join and leave the meeting. Each person independently picks two random times between 8 a.m. and 9 a.m. — not rounded to the nearest minute, mind you, but anytime within that range. They then join the meeting at the earlier time and leave the meeting at the later time.


What is the probability that at least one attendee is on the call with everyone else (i.e., the attendee’s time on the call overlaps with every other person’s time on the call)?

Extra credit: What is the probability that at least two attendees are on the call with everyone else?


1 Expert Answer

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Stanton D. answered • 05/31/20

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Stanton D.

You can model the data, with not much difficulty. I used LibreOffice. Run two columns as RND(), then 2 more columns to sort into MIN and MAX respectively (start and stop times for the call attendance, on the (0,1) interval). Then, roughly bunch the range, such as into 0.1 steps, and set up logical Boolean test: (end>range_beginning&start_range_end)[note: this seems contraintuitive, but it's easiest!)->numeric (0 or 1) output (in LibreCalc, that's the "NUM" conversion function) across the steps x range output area. The pattern of 1's shows if the rnd "call" is on anywhere in that step time. You may then either 1) sum the column values (shows the bulk attendance at that time interval). The bulk_on distribution is gently humped, with P~0.2 at the ends (but, that's with 0.1 step size; actually it must ->0 at the ends as a continuous function) and a max ~ 0.75 around t=0.5. or 2) product the column data (LibreOffice requires typing all cells for that, so you won't want to do it much!). Tests with just the first 4 callers showed a "hit" about 1/2 the trials, but with the first 10 callers, NO HITS in 11 trials. So I think that scales as the exponent(in #people) of P(bulk _not_on_the_call), which is astronomically small in your problem -- at best, ~0.75^330,000,000 or whatever. Do it by logs! And 2 people, would be the same, because the problem did specify that everyone else was on the call!! --Cheers, this was a fun problem to work on, --Mr. d.
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02/09/22

Stanton D.

You can model the data, with not much difficulty. I used LibreOffice. Run two columns as RND(), then 2 more columns to sort into MIN and MAX respectively (start and stop times for the call attendance, on the (0,1) interval). Then, roughly bunch the range, such as into 0.1 steps, and set up logical Boolean test: (end>range_beginning&start_range_end)[note: this seems contraintuitive, but it's easiest!)->numeric (0 or 1) output (in LibreCalc, that's the "NUM" conversion function) across the steps x range output area. The pattern of 1's shows if the rnd "call" is on anywhere in that step time. You may then either 1) sum the column values (shows the bulk attendance at that time interval). The bulk_on distribution is gently domed, with P~0.2 at the ends (but, that's with 0.1 step size; actually it must ->0 at the ends as a continuous function) and a max ~ 0.75 around t=0.5. or 2) product the column data (LibreOffice requires typing all cells for that, so you won't want to do it much!). Tests with just the first 4 callers showed a "hit" about 1/2 the trials, but with the first 10 callers, NO HITS in 11 trials. So I think that scales as the exponent(in #people) of P(bulk _not_on_the_call), which is astronomically small in your problem -- at best, ~0.75^330,000,000 or whatever. Do it by logs! And 2 people, would be the same, because the problem did specify that everyone else was on the call!! --Cheers, this was a fun problem to work on, --Mr. d.
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02/09/22

Stanton D.

By the way, sorry about Wyzant text modifications above. If you dont already know, it transforms most non-letter symbols into ampersand plus codes. Think you can decipher, except possibly where there is a Boolean-logical-and in the original text. I cant figure out what it did there.
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02/09/22

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