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April 2009 Q 8

  • Thread starter snerap@gmail.com
  • Start date
Suppose that x cats are currently infected. In part (i), we showed that the number of possible pairings of cats that could result in a new infection is x(10 - x). The total number of pairings of cats is 10C2 (ie 10 choose 2) = 45. So the probability that a contact is
between an infected cat and an uninfected cat is x(10 - x)/45.

When such a contact occurs, there is a 50% chance that it will lead to a new infection. So, given that there are currently x cats infected, the probability that a new infection occurs when two of the cats meet is x(10 - x)/90.

We are told that contacts between cats occur according to a Poisson process with parameter mu. To obtain the transition rates for the process X(t), to put in the generator matrix, we multiply the Poisson parameter by the probability that a contact leads to infection.
 
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