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Suppose 40% of the restaurants in a certain part of a town are in violation of the health code.
A health inspector randomly selects nine of the restaurants for inspection. (Round your answers to four decimal places.)
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(a) What is the probability that none of the restaurants are in violation of the health code?
- According to the condition, the probability that one single concrete restaurant does not violate the code is 60% = 0.6.
Hence, the probability that 9 of the restaurants do not violate the code is = 0.01 = 1%.
(b) What is the probability that one of the restaurants is in violation of the health code?
This probability is the complement to , which is the probability that all 9 of the restaurant do not violate the code
(see the solution (a)).
So, the answer is 1 - 0.01 = 0.99 = 99%.
(c) What is the probability that at least two of the restaurants are in violation of the health code?
The complement event is that none of the 9 restaurants violates the code or there is exactly one
among the 9 restaurants which does violate the code.
This probability of this complement event is .
Thus the probability under the question is = = 0.986 = 98.6%.