Question 392332
 2 TVs are defective, 6 non-defective

no. of ways to choose 3 TVs randomly from 8 = 8C3 = 56 


no. of ways to choose 3 non-defective TVs = 6C3 = 20


probability of no defective TVs in randomly selected 3 TVs =  20/56






....please write the question same as it is given on book or assignment.