discrete math
if you run a actual test your results will hardly ever exactly match the probabilty of a fair result.
but here is the probability of the events occurring. P(day 1)= 1=P(day 2).
let us label the (nine total) people 0 through 8
the first person infected we call patient zero, infects patient one (the second person infected) on day two, then the probability that one (the second person infected) will infect a new person is 7/8 there are only eight other people on the island besides themselves, and so on 6/8,5/8,4/8,3/8,2/8,1/8.
so on the third day there is a 7/8 = 0.875 chance of disease.
new infection on fourth day = 6/8 = 0.75
times previous 7/8 -> 7/8 * 6/8 = 42/64 =0.65625 chance of disease on day 4
new infection on fifth day 5/8 = 0.625
times all previous -> 42/64 * 5/8 = 210/512 =0.41 chance of disease on day 5
new infection on sixth day 4/8 = 0.50
times all previous -> 210/512 * 4/8 = 840/4096 =0.205 chance on day 6
new infection on seventh day 3/8 = 0.375
times all previous -> 840/4096 * 3/8 = 2520/32768 = 0.076904297 = P(day 7)
new infection on eighth day 2/8 = 0.25
times all previous new infections -> 5040/262144 = 0.019226074 = P(day 8).
new infection on ninth day 1/8 = 0.125
times all previous new infections -> 5040/2097152 = 0.002403259 =P(day 9).
it has been a while since you asked this question, but I thought it was a good thing to answer as a general solution to seeing approaches to probability word problems.
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