<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 probability that the first ball selected is green is $P_1 = \frac {3}{9} $. After selecting a green balls, there are now $2$ green and a total of $8$ balls. Now the probability of selecting a green balls is $P_2 = \frac {2}{8} $. Now there are a total of $7$ balls and a green ball. Now the probability is  $P_3=\frac {1}{7} $. Thus, $$P_{\text {req}} = P_1 \times P_2 \times P_3 = \frac {3}{9} \times \frac {2}{8} \times \frac {1}{7} = \frac {1}{84} $$ which is the same as that got using the hypergeometric distribution formula. Hope it helps. 
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