Defective Chips, Why not use nCk? Below is an example that I understand the solution to, but not why the problem should be approached using percentages versus n chose k. 

Example 1 from Khan Academy
A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way.
Suppose an inspector randomly selects 4 chips for an inspection.
Assuming the chips are independent, what is the probability that at least one of the selected chips is defective?
P(defective)=0.02 and P(not defective)=0.98, thus
P(No defective if 4 chosen) = 0.98^4 = 0.922
P(at least one defective) = 1 - 0.922 = 0.078 

What is wrong with using "nCk" (n choose k) to solve this problem? How would this problem change if the tested concept was meant to be "nCk".

Let $X$ be the number of defective chips among the four.
\begin{align}
\Pr(X\ge1) = {} & \Pr(X=1) + \Pr(X=2) + \Pr(X=3) + \Pr(X=4) \\[12pt]
= {} & {}_4C_1 \,\,0.98^1(1-0.98)^3 + {}_4C_2\,\,0.98^2(1-0.98)^2 \\[2pt]
& {} + {}_4C_3 \,\,0.98^3(1-0.98)^1 + {}_4C_4 \,\,0.98^4 \\[12pt]
= {} & 4\cdot0.98^1(1-0.98)^3 + 6\cdot0.98^2(1-0.98)^2 \\[2pt]
& {} + 4\cdot0.98^3(1-0.98) + 0.98^4.
\end{align}
That is perfectly valid. One can also proceed as follows:
$$
\Pr(X\ge 1) = 1 - \Pr(X=0) = 1 - 0.98^4.
$$
The second way is more efficient, especially if, for example, we had $60$ instead of $4.$
