A scientist inoculates several guinea pigs, one at a time, with the disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is 1/5, what is the probability that 8 guinea pigs are required?

Now let us assume that the Scientist injects a guinea pig with the germ and no disease is seen further, it can be estimated that some guinea pig has developed immunity and it is no point trying again with that particular animal. Again 8 animals are present. So less than or equal to that number might be utilized.

After three animals provide their response, no more injections are needed to be given. The events in which any two animals respond to the injection are known to be independent. Even in the case of 8 guinea pigs, no assurance can be made certainty that 3 would respond.

Then P(3 animals are used) = (1/5)^3 = P_3. P(<= 4 animals are used) = P_4 = P(3 responses in the first 3 animals) + P(2 response in the first 3 & 4th responds) = (1/5)^3 + 3 * (1/5)^2 * (4/5) * 1/5. More commonly, P(<= n animals are used) = P(3 responses in the first (n-1) attempts) + P(2 responses in the first (n-1) attempts & nth animal responds).

P_n = P_(n-1) + (n-1 * (n-2) * (1/5)^2 * (4/5)^(n-3) /2 Since P_3 is known, P_8 can be calculated (maybe using a spreadsheet).

If you mean exactly 8 animals are used, It gets easier because only the last expression in P_n is needed. This kind of problem is best understood with a state transition diagram.

To know more about probability:

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