an urn contains six ball of each of the three colors: red, blue, and green. An urn contains six balls of each of the three colors: red, blue, green. Find the expected number of different colors obtained when three balls are drawn:
a. with replacement;
b. without replacement.
The correct answers are:
a: 19/9
b: 2.1912
I just don't know how to get to get to the answers.
Thanks for any help
 A: We can also use recurrences, where A,B,C are the number of balls of three colors:
a)
\begin{align*}
  f(a,b,c) = \left\{\begin{matrix}
  3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)&   \text{if }A+B+C-(a+b+c) = 3\\ 
  \dfrac{a\cdot f(a-1,b,c)+b\cdot f(a,b-1,c)+c\cdot f(a,b,c-1)}{a+b+c} &  \text{otherwise}
\end{matrix}\right.
\end{align*}
b)
\begin{align*}
  f(a,b,c) = \left\{\begin{matrix}
  3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)&   \text{if }A+B+C-(a+b+c) = 3\\ 
  \dfrac{A\cdot f(a-1,b,c)+B\cdot f(a,b-1,c)+C\cdot f(a,b,c-1)}{A+B+C} &  \text{otherwise}
\end{matrix}\right.
\end{align*}
A: The count, $X,Y$, of different colours drawn with and without replacement (respectively) will have the support: $\{1, 2, 3\}$.   Find the probability of each, for the two experiments, then use:
$$\begin{align}\mathsf E(X) & = \sum_{k=1}^3 k\;\mathsf P(X{=}k) \\[1ex] & = \mathsf P(X{=}1)+2\,\mathsf P(X{=}2)+3\,\mathsf P(X{=}3) \\[1ex] & = 2 -\mathsf P(X=1)+\mathsf P(X=3) \\[2ex]\mathsf E(Y) & = 2 -\mathsf P(Y=1)+\mathsf P(Y=3) \end{align}$$
$\mathsf P(X=1), \mathsf P(Y=1)$ are the probabilities that all three balls are the same colour.
$\mathsf P(X=3), \mathsf P(Y=3)$ are the probabilities that all three balls are of distinct colours.
A: Let us take a card deck that contains only three colors : spade, diamond and heart, and suits from 1 to 6.
1) Define a poker hand as a hand of three cards. There are several types of hands, full color, pair and tricolor. Let count the hands without replacement :
tricolor : $ {6 \choose 1} {6 \choose 1} {6 \choose 1} = 216 $
pair : $ {3 \choose 1} {2 \choose 1} {6 \choose 1} {6 \choose 2} = 540 $
full color : $ {3 \choose 1} {6 \choose 3}= 60 $
their sum is $ {18 \choose 3} = 816 $ , the average # of colors is ${ 3×216 + 2×540 + 1×60 \over 816 }\approx 2.1912   $
2) When replacing, the G.F. is  $(x+y+z)^3$ 
The needed coefficients satisfy 6 + 18 + 3 = 27;
The average # of colors is ${ 3×6 + 2×18 + 1×3 \over 27 } = {19 \over 9}   $
