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
$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$
 A: 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. 
A: The answer is given by:
$$
P = \frac{3}{9} \cdot \frac{2}{8}\cdot \frac{1}{7} = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{7}  = \frac{1}{84}
$$
A: a) Your first answer is correct.
b) All 3 are green without replacement 
= $\dfrac39 \cdot  \dfrac28 \cdot  \dfrac17$ =  $\dfrac1{84}$
c) 2 balls are green out of 3 without replacement.
= $ \left(\dfrac{3}{9} \cdot  \dfrac{2}{8} \cdot \dfrac{6}{7} \right ) + \left(\dfrac{3}{9} \cdot  \dfrac{6}{8} \cdot \dfrac{2}{7} \right ) + \left(\dfrac{6}{9} \cdot  \dfrac{3}{8} \cdot \dfrac{2}{7} \right )$ 
= $3 \cdot \left(\dfrac{3 \cdot 2 \cdot 6}{9 \cdot 8 \cdot 7}\right)$ = $\dfrac{3}{14}$
Or
= $ \binom{3}{2} \cdot \left(\frac{3}{9} \cdot  \dfrac{2}{8} \cdot \dfrac{6}{7} \right )$  =  $\frac{3}{14}$
Here $\binom{3}{2}$ because 2 greens can come to any place out of 3.
