Question

a large shipment of video streaming devices is accepted upon delivery if an inspection of 20 randomly selected video streaming devices yields no more than 1 defective item.

Find the probability that the shipment is accepted if 2% of the total shipment is defective.

Find the probability that the shipment is accepted if 10% of the total shipment is defective.

(Solve without using excel)

Answer 1

    Let X be the number of defective devices in the sample of 20                   
   n = 20   (sample size of devices)              
   PDF of X is given by                  

   where p is the probability of selecting a defective device                  
   and q = 1 - p                  
   For accepting the shipment, the number of defectives should not be more than 1                  
   Thus probability of accepting a shipment = P(X ≤ 1)                  
                      
Case 1)   2% of the total shipment is defective                  
   Thus,                   
   p = 0.02       (probability of selecting a defective device from the total devices)          
   q = 1 - p = 1 - 0.02 = 0.98                  
   Thus, X ~ Binomial distribution with n = 20 and p = 0.02                  
   Probability of accepting a shipment = P(X ≤ 1)                  
   P(X ≤ 1) = P(X = 0) + P(X = 1)                  
             
                   = 1*1*(0.98)20 + 20*0.02*(0.98)19                  
                   = 0.6676 + 0.2725                  
                   = 0.9401                  
   Probability that the shipment is accepted if 2% of the total shipment is defective = 0.9401                  
                      

Case 2)   10% of the total shipment is defective                  
   Thus,                   
   p = 0.10       (probability of selecting a defective device from the total devices)          
   q = 1 - p = 1 - 0.10 = 0.90                  
   Thus, X ~ Binomial distribution with n = 20 and p = 0.10                  
   Probability of accepting a shipment = P(X ≤ 1)                  
   P(X ≤ 1) = P(X = 0) + P(X = 1)                  
         
                   = 1*1*(0.90)20 + 20*0.10*(0.90)19                  
                   = 0.121577 + 0.27017                  
                   = 0.391747                  
   Probability that the shipment is accepted if 10% of the total shipment is defective = 0.391747