Why taking combinations works but conventional approach does not in this case? The question is 
Let two items be chosen from a lot containing 12 items of which 4 are defective. Find the probability that atleast one item is defective.
Approach 1:
$$
\frac{4}{12}\times\frac{8}{11}+\frac{4}{12}\times\frac{3}{11} = \frac{1}{3}
$$
First term represents the case in which one item is defective and second is not defective.
Here $\frac{4}{12}$ is probability choosing one item from $4$ defective items out of $12$ items.
As a result, now there are $11$ items are there to choose from. $\frac{8}{11}$ is probability of choosing one from $8$ non-defective item out of $11$ items.
Second term represents the case in which both items are defective. $\frac{4}{12}$ is probability of choosing one item from $4$ defective items and $\frac{3}{11}$ is probability of choosing another item from $3$ defective items.
Approach 2:
Total number of ways of choosing $2$ items out of $12$ $= \binom{12}{2} = 66$
Total number of ways of choosing $2$ non-defective items out of 8 $= \binom{8}{2} = 28$
Let $X$ denote number of defective items chosen.
Therefore,
$$
P(X = 0) = \frac{14}{33}
$$
Probability of choosing atleast one defective item is denoted by $P(X \ge 1)$. Therefore,
$$
\begin{split}
P(X \ge 1) &= 1 - P(X = 0) \\
&= 1 - \frac{14}{33} \\
&= \frac{19}{33}
\end{split}
$$
Why the first approach gives a wrong answer and why second approach works?

In the first approach you have to count the first term twice, representing the fact that you can either pull a working item and then a defective one or the other way around. This gives
$$2×\frac4{12}×\frac8{11}+\frac4{12}×\frac3{11}=\frac{19}{33}$$
