Question 1118735
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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  {{{0.6^9}}} = 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 {{{0.6^9}}}, 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  {{{0.6^9 + C[9]^1*0.6*0.4^8}}}.

     Thus the probability under the question is  {{{1 - 0.6^9 - C[9]^1*0.6*0.4^8}}} = {{{1 - 0.6^9 - 9*0.6*0.4^8}}} = 0.986 = 98.6%.
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