A box has five items, three good and two defective. three items are selected at random without replacement. let x be the number of defective items selected. find the probability mass function of x.

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}

Therefore the 
PMF:
{(0, 0.1),(1, 0.6),(2, 0.3)}