Final answer:
The probability of randomly meeting either a woman or an American in a group consisting of 110 individuals with a distribution of German men, German women, American men, and American women is 68.18%.
Step-by-step explanation:
The question is asking for the probability of meeting either a woman or an American when randomly selecting a person from a group. The group is composed of 35 German men, 25 German women, 20 American men, and 30 American women. To find this probability, we can add the probabilities of meeting a woman (regardless of nationality) and meeting an American (regardless of gender), and then subtract the probability of meeting an American woman, since this group has been counted twice in our previous addition.
The total number of people in the group is 35 (German men) + 25 (German women) + 20 (American men) + 30 (American women) = 110. The number of women is 25 (German women) plus 30 (American women) equaling 55. The number of Americans is 20 (American men) plus 30 (American women) equaling 50. The number of American women, which we count once, is 30.
Thus, the probability of meeting either a woman or an American is given by:
P(woman or American) = (number of women + number of Americans - number of American women) / total number of people
P(woman or American) = (55 + 50 - 30) / 110 = 75 / 110 = 15/22 ≈ 0.6818
Therefore, the probability of randomly meeting either a woman or an American in this group is approximately 0.6818 or 68.18%.