G Of 23 fast food restaurants in the city, a total of 5 were in violation of sanitary standards, a total of 6 were in violation of safety standards, and 3 were in violation of both. If a fast food restaurant is chosen at random from these 23, what is the probability that it is in compliance of both safety and sanitary standards?

Answer:65.22% probability that it is in compliance of both safety and sanitary standardsStep-by-step explanation:A probability is the number of desired outcomes divided by the number of total outcomes.Number of restaurant in compliance of both safety and sanitary standards.First I will find those not in compliance, using a Venn's set.Set A: Violation of sanitary standards.Set B: Violation of safety standards.5 were in violation of sanitary standards, a total of 6 were in violation of safety standards, and 3 were in violation of both.This means that [tex]A = 5, B = 6, (A \cap B) = 3[/tex]At least one:[tex]A \cup B = A + B - (A \cap B)[/tex]So[tex]A \cup B = 5 + 6 - 3 = 8[/tex]8 are in violation of at least one of these standards.So 23-8 = 15 are in compliance of both safety and sanitary standards.What is the probability that it is in compliance of both safety and sanitary standards?15 out of 2315/23 = 0.652265.22% probability that it is in compliance of both safety and sanitary standards