Answer:
65.22% probability that it is in compliance of both safety and sanitary standards
Step-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 23
15/23 = 0.6522
65.22% probability that it is in compliance of both safety and sanitary standards