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Topic: [ap-stat] RE: Spread of an epidemic
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pat ballew

Posts: 334
Registered: 12/3/04
[ap-stat] RE: Spread of an epidemic
Posted: Nov 1, 2000 7:14 PM
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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 (1-8) 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 non-immune monks out of the seven other monks ... hence

I think from here the method is clear if I am really understanding the

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

Pat Ballew,
Misawa, Jp

"Statistics means never having to say you're certain."

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-----Original Message-----
From: Mark Fountain [mailto://]
Sent: Thursday, November 02, 2000 7:25 AM
To: AP Statistics
Subject: [ap-stat] 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 24-hour virus. While contagious, the contagious monk randomly
visit one of the other monks. Twenty-four 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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