Suppose Group A makes up an estimated 2% of the population and Group B makes up an estimated 98% of the population (Group B consists of everyone who is not in Group A). However, Group A accounted for 80% of the estimated new virus X infections.
Question: How many times more likely are the members of Group A to be diagnosed with virus X compared to the members of Group B?
I tried comparing the ratios: $(98/2) \div (20/80) = 49 \times 4 = 196$, but I do not know if this is the correct approach, and I do not really know how to explain my reasoning.
Any help would be appreciated.
