Binomial vs Probability 
A shipment of 12 microwave ovens contains 3 defective units. A vending company has ordered 4 of these units, and because all are packaged identically, the selection will be at random. What is the probability that at least 2 units are good?

There are two ways to solve this problem:


*

*Using the binomial distribution: 
\begin{align*}
P(\text{defective}) & = \frac{3}{12} = \frac{1}{4}\\
P(\text{good}) & = 1 - \frac{1}{4} = \frac{3}{4}\\
P(2~\text {good}) & = \binom{4}{2} \times 0.75^{2} \times 0.25^{2} = 0.211\\
P(3~\text{good}) & = \binom{4}{3}\times 0.75^{3} \times 0.25 = 0.422\\
P(4~\text{good}) & = 0.75^{4} = 0.316
\end{align*}
The probability that at least 2 units are good is given by:
$$P(2~\text{good}) + P(3~\text{good}) + P(4~\text{good}) = 0.211 + 0.422 + 0.316 = 0.949$$

*Using combinations:
$$\frac{\dbinom{9}{2}\dbinom{3}{2} + \dbinom{9}{3}\dbinom{3}{1} + \dbinom{9}{4}}{\dbinom{12}{4}} = \frac{486}{495} = \frac{54}{55} = 0.9818$$ 
Why are these two answers different?

You cannot use the binomial distribution, as this is an experiment without replacement: once you pick a defective oven, there are only two defective ovens left. Using combinations, the probability of selecting at least two good units indeed equals:
$$\frac{{9 \choose 4}{3 \choose 0} + {9 \choose 3}{3 \choose 1} + {9 \choose 2}{3 \choose 2}}{12 \choose 4} \approx 0.982$$
