Question 320334
a) all are defective. 
This means selecting four defectives in a row.  The first one being defective is 4/24, the second being defective is 3/23 the third one being defective is 2/22 and the fourth one being defective is 1/21, therefore P(all defective) = P(1st defective)*P(2nd defective)*P(3rd defectie)*P(4th defective) = 4/24*3/23*2/22*1/21=24/255024 = 0.000094

b)one being defective
This means 1 defective and 3 non-defective =4C1* P(D)*P(N)*P(N)*P(N)*P(N) since there are 4C1 ways to arrange that "one" defective. 4*4/24*20/23*19/22*18/21=109440/255024=0.429