3 ms·
Probability of 1 town of 10^5 people having 4 people disease with 1 in 10^-6: (10^5 choose 4)x(10^-6)^4x(1-10^-6)^9996 ~= 3.77x10^-6 Assuming independence betw
by hervature 8y ago
Probability of 1 town of 10^5 people having 4 people disease with 1 in 10^-6: (10^5 choose 4)x(10^-6)^4x(1-10^-6)^9996 ~= 3.77x10^-6
Assuming independence between diseases and towns (not a terrible assumption but definitely not true) we have:
2000x200x3.77x10^-6 ~= 1.5
Which agrees with your simulation [1].
[1] My slow brain and calculator took 2 minutes to calculate this ;)
EDIT: Changed asterisks to x to avoid formatting.
EDIT 2: Here are the numbers for all the cases you have:
0: 361934.8
1: 36913.8
2: 1809.5
3: 60.3
4: 1.5