Ascientist inoculates mice, one at a time, with a disease germ 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?

There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).

13=7.8514=3.5feet15=10.99feet16=2& 317=418=about 9ft19=7.8520=4in

step-by-step explanation: