<s>[INST] Bag marbles with three colors and probability A bag contains one red, two blue, three green, and four yellow balls.  A sample of three balls is taken without replacement.  Let $Y$ be the number of yellow balls in the sample. Find the probability of $Y=0$, $Y=1$, $Y=2$ $Y=3$

Attempt 1
all three are yellow would be 
$$ \frac{\binom{4}{3}}{\binom{10}{3}}=\frac{4}{120}=\frac{1}{30}  $$
two are yellow combinations would be
 yy red
$$ \binom{4}{2}\binom{1}{1} =6$$
yy green
$$ \binom{4}{2}\binom{3}{1}=18 $$
yy blue 
$$ \binom{4}{2}\binom{2}{1} = 12$$
adding them together and dividing getting $P(Y=2)=.3$
for one yellow
it would be 
combinations for Yellow red blue
combinations  for Yellow red green
combinations  for Yellow b   b
combinations  for Yellow b   g
combinations  for Yellow g   g
adding them all together and dividing by $120$
for no yellows
just keep going???Better algorithm?
 [/INST] We have four yellow balls and $1 + 2 + 3 = 6$ balls that are not yellow.  The number of ways we can select exactly $k$ yellow balls and $3 - k$ balls that are not yellow is 
$$\binom{4}{k}\binom{6}{3 - k}$$
Since there are $\binom{10}{3}$ possible selections of three of the ten balls, the desired probabilities are 
\begin{align*}
P(Y = 0) & = \frac{\dbinom{4}{0}\dbinom{6}{3}}{\dbinom{10}{3}}\\
P(Y = 1) & = \frac{\dbinom{4}{1}\dbinom{6}{2}}{\dbinom{10}{3}}\\
P(Y = 2) & = \frac{\dbinom{4}{2}\dbinom{6}{1}}{\dbinom{10}{3}}\\
P(Y = 3) & = \frac{\dbinom{4}{3}\dbinom{6}{0}}{\dbinom{10}{3}}
\end{align*}
</s>