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Topic:
[apstat] RE: Spread of an epidemic
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[apstat] RE: Spread of an epidemic
Posted:
Nov 1, 2000 7:14 PM


If I understand the question, the only way the virus will survive to infect everyone is if the random selection never repeats a monk. thus the probability is just the probability that you pick 8 numbers in a row at random from (18) and never repeat a value...
The initial monk is monk zero, and we assume he WILL infect someone so the probability of one transmission is P(1)=1 The probability of a second transmission is the chance that monk one will visit one of the six monks other than monk zero, hence p(2)= 6/7...
The probability of a third transmission is the chance that monk one infected someone (monk 2) times the probability that he found one of the five remaining nonimmune monks out of the seven other monks ... hence P(3)=6/7*5/7
I think from here the method is clear if I am really understanding the problem...
The probability then, that the epidemic ends after one infection is 1/7, after two infections is 6/7*2/7,  P(2) times NOT a third... after three infections is 6/7 * 5/7 * 3/7 .... and the probability that it infects everyone is 6/7*5/7*4/7*4/7*2/7*1/7
I think I did that right, but as soon as I push send I'll find something.... Hmmmm.... well here goes..
Pat Ballew, Misawa, Jp
"Statistics means never having to say you're certain."
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Original Message From: Mark Fountain [mailto://74023.221@compuserve.com] Sent: Thursday, November 02, 2000 7:25 AM To: AP Statistics Subject: [apstat] Spread of an epidemic
Hi, folks.
This is a question about YMM Special Problem 5A, The Spread of an epidemic. The setup is like this. There are 8 monks and one is infected with a 100% contagious 24hour virus. While contagious, the contagious monk randomly visit one of the other monks. Twentyfour hours later he is immune but the new carrier visits another randomly chosen monk. If that monk is already immune, the virus will die out; if not, the cycle continues. Will the virus die out before everyone is infected?
The task is to setup a similation and repeat it several times. However, one group is really hung up on the theoretical probability of this happening. Here's my best guess so far. There are 8! different orders that would infect every monk. There are 8^8 different ways to choose 8 random numbers since we have to allow replacement. So the probability is 8!/(8^8) = .002ish. What think you? I'm really not very confident with probability questions.
Thanks for your input. Sue Fountain
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