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Whats the probability of not getting a disease today where. t = transmition rate given contact with infected individual r = % of population infected c = number of contacts today

is it A) (1-rt)^c i.e. I meet c people , for each person the chance is transmitionRate * chance they have disease.

or

B) (1- t)^cr i.e. I'm likely to meet cr infected people , each meeting has a probability of t to transfer the disease.

Both seem valid, have similar but not identical results. e.g. if r=0.01 , c =100, t =0.5. the results are A~0.6 B=0.5. B seems accurate if I imagine the entire population = 100, then the actual result = 0.5.

Whats the difference/breakdown in the logic ?

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    $\begingroup$ I find it helpful to consider which axioms of probability can be used to justify formulas like (A) or (B). In this case one of them can be justified while the other cannot. The very language you use hints at this: you seem to be hedging a little by writing "likely to meet," rather than claiming explicitly that this really is a chance you are computing. That suggests to me that you already have a pretty good idea what the answer is :-). $\endgroup$ – whuber Feb 2 '15 at 22:11
  • $\begingroup$ I might have been loose with my language with "likely to meet" , Im genuinely stumped as to the correct. Happy to read of the axiom one is breaking? My guess is that (A) is correct , but B gives the correct answer for a fully specified population. This is just for my own interest so full answer welecome. $\endgroup$ – Hugh Feb 2 '15 at 22:51
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For this to be answerable with the given information, we need to assume some independence. So, assume that people that you can meet get the disease independent of each other. First let us find the probability for you getting the disease in one contact: $$ P(\text{you get disease in one contact})=P(\text{the contact has the disease})\cdot P(\text{disease get transmitted})= r\cdot t $$ For you to not get the disease in $c$ contact, you must avoid the disease in each and every contact, so $$ P(\text{not getting disease in $c$ contacts}) = (1-rt)^c $$ (here we are using the independence assumption). This is your answer A).

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