<s>[INST] Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green? 
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls are green?
Is it $\dfrac{3}{9}\times\left\{\dfrac{2}{8}+\dfrac{3}{8}\right\}\times\left\{\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}\right\}=\dfrac{90}{504}$?
But in hypergeometric distribution formula,  it is
$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$
 [/INST] The answer is given by:
$$
P = \frac{3}{9} \cdot \frac{2}{8}\cdot \frac{1}{7} = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{7}  = \frac{1}{84}
$$
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