a health instructor surveyed all her students, and tallied the results. the first survey question asked, \do…

a health instructor surveyed all her students, and tallied the results. the first survey question asked, \do you adhere to a vegetarian diet or omnivorous diet?\ the second question asked, \approximately how often do you exercise?\ what is the probability that a randomly selected student exercises weekly given that the student adheres to an omnivorous diet? simplify any fractions.

a health instructor surveyed all her students, and tallied the results. the first survey question asked, \do you adhere to a vegetarian diet or omnivorous diet?\ the second question asked, \approximately how often do you exercise?\ what is the probability that a randomly selected student exercises weekly given that the student adheres to an omnivorous diet? simplify any fractions.

Answer

Explanation:

Step1: Identify relevant numbers

We want the probability of weekly - exercise given omnivorous diet. The number of students with an omnivorous diet and weekly exercise is 1, and the total number of students with an omnivorous diet is (3 + 1=4).

Step2: Apply conditional - probability formula

The formula for conditional probability (P(A|B)=\frac{P(A\cap B)}{P(B)}). In terms of frequencies, if (A) is the event of weekly exercise and (B) is the event of an omnivorous diet, then (P(A|B)=\frac{n(A\cap B)}{n(B)}), where (n(A\cap B)) is the number of elements in (A\cap B) and (n(B)) is the number of elements in (B). So the probability is (\frac{1}{3 + 1}).

Answer:

(\frac{1}{4})