SOLUTION: 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 y

(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%.