Question

select the correct answer from each drop - down menu. a group of people were surveyed, and the data about their age and whether they have an iron deficiency was recorded in the two - way frequency table. iron deficiency|age| |---|---|---|---| | |less than 20 years|20 - 30 years|more than 30 years|total| |yes|41|37|24|102| |no|109|43|46|198| |total|150|80|70|300| based on the data in the table, the probability of randomly choosing a person who has iron deficiency and is less than 20 years old is. the probability of randomly choosing a person who does not have iron deficiency and is 20 years old or more is.

Explanation:

Step1: Recall probability formula

The probability formula is $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

Step2: Calculate probability of person with iron - deficiency and less than 20 years old

The number of people with iron - deficiency and less than 20 years old is 41. The total number of people surveyed is 300. So the probability $P_1=\frac{41}{300}$.

Step3: Calculate number of people without iron - deficiency and 20 years old or more

The number of people without iron - deficiency and 20 - 30 years old is 43, and the number of people without iron - deficiency and more than 30 years old is 46. So the number of favorable outcomes is $43 + 46=89$. The total number of people surveyed is 300. So the probability $P_2=\frac{89}{300}$.

Answer:

The probability of randomly choosing a person who has iron deficiency and is less than 20 years old is $\frac{41}{300}$. The probability of randomly choosing a person who does not have iron deficiency and is 20 years old or more is $\frac{89}{300}$.