Of 18 fast food restaurants in a city,7 are in violation of sanitary standards, 8 are in violation of safety standards, and 4 are in violation of both.If a fast food restaurant is chosen at random, what is the probability that it is in compliance with both safety and sanitary standards? The answer is 7/18 and I have no idea how. Solution (AUB)=(A)+(B)-(AnB) where A= sanitary standared Violation , B=saftey stndrd violation AUB=7+8-4=11 now The restaurants which will be compliance with both A and B will be denoted by 18- AUB=18-11=7 so the probablity that the random selected restaurant is in compliance with both safety and sanitary standards is= (18-AUB)/18=7/18.

1. Of 18 fast food restaurants in a city,7 are in violation of sanitary standards, 8 are in violation of
safety standards, and 4 are in violation of both.If a fast food restaurant is chosen at random, what
is the probability that it is in compliance with both safety and sanitary standards?
The answer is 7/18 and I have no idea how.
Solution
(AUB)=(A)+(B)-(AnB) where A= sanitary standared Violation , B=saftey stndrd violation
AUB=7+8-4=11
now The restaurants which will be compliance with both A and B will be denoted by 18-
AUB=18-11=7
so the probablity that the random selected restaurant is in compliance with both safety and
sanitary standards is=
(18-AUB)/18=7/18