This about probability someone to help me on this !: A scientist inoculates several mice,one at time,until he finds 2 that have contracted the disease.If the probability of contracting the disease is 1/6 , what is the probability that 8 mice are required?

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P( disease) = 1/6 P(no disease) = 5/6 then is a binomial probability problem 2 diseased 8C2 (1/6)^6(5/6)^2 I think... ages since I did this stuff

you have disease and no disease mixed up no?

I think the power 6 is on 5/6,and power 2 is on 1/6 because our combination contain 6 mice not contracted and 2 mice contracted , I think also that final element of our combination must be contracted mouse ; my suggestion is 7C1 (1/6)^2 (5/6)^6; I need a comment on this

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