If 3% of a concrete mix produced by a company are defective, determine the probabilities that in a sample of 80 concrete cubes...

Question

a) Two and more than two will be defective.

Answer 1

BInomial p=0.8 N=80

 

1 - [ Prob(X= 1) + Prob(X=0)] 

 

 

Prob(X=1) = ( 80 choose 1) (0.8)^79 (0.2)^1 = 80*(0.8)^79*0.2  which is essentially zero 2.5336942 x 10 ^-7

 

Prob(X=0) = (80 choose 0) (0.8)^80 * (0.2)^0 = 1 * 0.8^80 * 1 = 0.8^80 = 1.767 x 10^-8

 

The probability will be greater than 99%