This problem can be solved from first principles, case by case. However, it can be solved systematically using the hypergeometric distribution, based on the characteristics of the problem: - known number of defective and non-defective items. - no replacement - known number of items selected.
Let a=number of defective items selected A=total number of defective items b=number of non-defective items selected B=total number of non-defective items Then P(a,b)=C(A,a)C(B,b)/C(A+B,a+b) where C(n,r)=combination of r items selected from n, A+B=total number of items a+b=number of items selected
Given: A=2 B=3 a+b=3 PMF: P(0,3)=C(2,0)C(3,3)/C(5,3)=1*1/10=1/10 P(1,2)=C(2,1)C(3,2)/C(5,3)=2*3/10=6/10 P(2,0)=C(2,2)C(3,1)/C(5,3)=1*3/10=3/10 Check: (1+6+3)/10=1 ok note: there are only two defectives, so the possible values of x are {0,1,2}
